Moving Charges and Magnetism

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Summary

Chapter Summary: Moving Charges and Magnetism

Key Concepts

Electricity and Magnetism Relationship: Realized in 1820 by Hans Christian Oersted through experiments showing current affects magnetic compass.
Magnetic Field: Produced by moving charges or currents.

Important Formulas

Current Enclosed:

$I_e = I \left( \frac{\pi r^2}{\pi a^2} \right) = \frac{Ir^2}{a^2}$
- Indicates current enclosed by a circle of radius r.
Magnetic Field Using Ampere’s Law:

$B(2\pi r) = \mu_0 \frac{Ir^2}{a^2}$
- Simplifies to:
  $B = \left( \frac{\mu_0 I}{2\pi a^2} \right) r$
- Implies:
  $B \propto r \quad (r < a)$
Magnetic Field from a Long Straight Wire:

$B = \frac{\mu_0 I}{2\pi R}$
- Field lines are concentric circles around the wire.
Magnetic Field Inside a Solenoid:

$B = \mu_0 n I$
- n is the number of turns per unit length.
Magnetic Moment:

$m = N/A$
- Direction given by right-hand thumb rule.
Torque on a Loop in a Magnetic Field:

$\tau = m \times B$

Learning Objectives

Understand the relationship between electricity and magnetism.
Apply Ampere's law to calculate magnetic fields.
Analyze the behavior of charged particles in magnetic fields.
Calculate magnetic forces and torques on current-carrying loops.

Common Mistakes & Exam Tips

Confusing Right-Hand and Left-Hand Rules: Ensure to apply the correct rule based on the charge type (positive or negative).
Forgetting Units: Always include units in calculations, especially for magnetic fields (Telsa).
Neglecting Direction: Remember that magnetic forces are vector quantities and have direction.

Important Diagrams

Magnetic Field Lines: Illustrate how magnetic field lines form closed loops and are affected by current direction.
Lorentz Force: Diagrams showing the force on charges moving in magnetic fields, indicating the use of right-hand and left-hand rules for direction determination.

Learning Objectives

Understand the relationship between electricity and magnetism.
Explain the concept of magnetic field lines and their properties.
Apply Ampere's Circuital Law to calculate magnetic fields.
Derive the expression for the magnetic field due to a long straight wire.
Analyze the behavior of magnetic fields in solenoids and coils.
Calculate the magnetic moment of a current-carrying loop.
Describe the operation of a moving coil galvanometer and its conversion to an ammeter or voltmeter.
Solve problems involving the Lorentz force and its implications in electromagnetic theory.
Evaluate the effects of parallel and anti-parallel currents on magnetic interactions.

Detailed Notes

Chapter Notes: Moving Charges and Magnetism

1. Introduction

Electricity and Magnetism have been known for over 2000 years.
In 1820, Hans Christian Oersted discovered that a current in a wire deflects a magnetic compass needle, indicating a relationship between electricity and magnetism.

2. Physical Quantities

Physical Quantity	Symbol	Nature	Dimensions	Units	Remarks
Permeability of free space	µ₀	Scalar	[MLT⁻²A⁻²]	Tm A⁻¹	4π X 10⁻⁷ Tm A⁻¹
Magnetic Field	B	Vector	[M T⁻²A⁻¹]	T (telsa)
Magnetic Moment	m	Vector	[L²A]	A m² or J/T
Torsion Constant	k	Scalar	[M L²T⁻²]	N m rad⁻¹	Appears in MCG

3. Key Concepts

3.1 Electrostatic and Magnetic Fields

Electrostatic field lines originate at positive charges and terminate at negative charges or fade at infinity.
Magnetic field lines always form closed loops.

3.2 Lorentz Force

The expression for the Lorentz force is given by:

$F = q (v imes B + E)$
This force is dependent on the velocity of the charged particle.

3.3 Ampere's Circuital Law

Ampere's law relates the magnetic field around a closed loop to the current passing through the loop:

$B imes L = µ₀ I_e$
Where:
- L is the length of the loop
- I_e is the net current enclosed by the closed circuit.

3.4 Magnetic Field from a Long Straight Wire

The magnetic field at a distance R from a long straight wire carrying a current I is given by:

$B = rac{µ₀ I}{2πR}$

3.5 Magnetic Field Inside a Solenoid

The magnetic field B inside a long solenoid carrying a current I is:

$B = µ₀ n I$
Where n is the number of turns per unit length.

4. Important Relationships

Parallel currents attract, while anti-parallel currents repel.
The magnetic moment m of a planar loop carrying a current I with N turns and area A is:

$m = N imes A$

5. Exercises

Exercise 4.1: Calculate the magnetic field at the center of a circular coil with 100 turns and radius 8.0 cm carrying a current of 0.40 A.
Exercise 4.2: Determine the magnetic field at a point 20 cm from a long straight wire carrying a current of 35 A.
Exercise 4.3: Find the magnitude and direction of the magnetic field at a point 2.5 m east of a wire carrying a current of 50 A.

6. Points to Ponder

The relationship between electricity and magnetism is fundamental in understanding electromagnetic phenomena.
The behavior of magnetic fields around current-carrying wires can be visualized using compass needles and iron filings.
The principles of electromagnetism have led to significant technological advancements in the 20th century.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding the Relationship Between Electricity and Magnetism: Students often confuse the principles governing electricity and magnetism. Remember that they are linked phenomena, as demonstrated by Oersted's experiment.
Ignoring the Right-Hand Rule: When applying Ampere's Circuital Law, failing to correctly apply the right-hand rule can lead to incorrect signs for the current.
Forgetting the Conditions for Steady Currents: The laws discussed in this chapter apply only to steady currents. Be cautious when dealing with varying currents, as Newton's third law may not hold without considering the electromagnetic field's momentum.
Confusing Magnetic Field Lines: Unlike electric field lines, which originate and terminate at charges, magnetic field lines always form closed loops. This fundamental difference is often overlooked.

Tips for Success

Practice Using the Right-Hand Rule: Regularly practice applying the right-hand rule to determine the direction of magnetic fields and forces. This will help solidify your understanding.
Visualize Magnetic Fields: Use diagrams to visualize how magnetic fields are generated by current-carrying wires. Sketching can help reinforce concepts.
Review Key Formulas: Familiarize yourself with key formulas related to magnetic fields, such as those for a long straight wire and solenoids. Make a list of these formulas for quick reference.
Work Through Examples: Solve example problems similar to those in your exercises. This will help you understand the application of concepts and formulas in different scenarios.
Understand the Concept of Magnetic Moment: Make sure you grasp the concept of magnetic moment and how it relates to current loops, as this is crucial for understanding torque in magnetic fields.

Practice & Assessment

Multiple Choice Questions

A current in a wire can deflect a nearby magnetic compass needle.

Electricity and magnetism are unrelated phenomena.

Magnetic fields can be created only by permanent magnets.

Electric fields are unaffected by magnetic fields.

Correct Answer: A

Solution:

Oersted discovered that a current in a straight wire causes a deflection in a nearby magnetic compass needle, indicating a relationship between electricity and magnetism.

They repel each other.

They attract each other.

They have no effect on each other.

They create a circular magnetic field.

Correct Answer: B

Solution:

Parallel currents attract each other due to the magnetic fields they produce.

It converts the galvanometer into a voltmeter

It increases the sensitivity of the galvanometer

It converts the galvanometer into an ammeter

It decreases the resistance of the galvanometer

Correct Answer: C

Solution:

Introducing a shunt resistance in parallel with a moving coil galvanometer converts it into an ammeter by allowing most of the current to bypass the galvanometer.

It determines the sensitivity of the galvanometer

It balances the magnetic torque

It measures the current directly

It provides electrical resistance

Correct Answer: B

Solution:

The torsion constant

k

in a moving coil galvanometer provides a counter-torque that balances the magnetic torque.

The magnetic field strength increases.

The magnetic field strength decreases.

The magnetic field strength remains the same.

The magnetic field direction reverses.

Correct Answer: A

Solution:

The magnetic field inside a long solenoid is given by

B = \mu_0 n I

, where

n

is the number of turns per unit length. Increasing

n

increases the magnetic field strength.

B = \mu_0 n I

B = \frac{\mu_0 n I}{2}

B = \frac{\mu_0 I}{n}

B = \mu_0 n^2 I

Correct Answer: A

Solution:

The magnetic field inside a long solenoid is given by

B = \mu_0 n I

, where

\mu_0

is the permeability of free space,

n

is the number of turns per unit length, and

I

is the current.

Upwards

Downwards

Towards the wire

Away from the wire

Correct Answer: A

Solution:

According to the right-hand rule, if you grasp the wire with your right hand with your thumb pointing in the direction of the current (north to south), your fingers will curl in the direction of the magnetic field. At a point east of the wire, this direction is upwards.

The field is zero.

The field is uniform and strong.

The field is weak and varies along the axis.

The field is perpendicular to the axis.

Correct Answer: B

Solution:

Inside a long solenoid, the magnetic field is uniform, strong, and directed along the axis of the solenoid.

Parallel to the current.

Perpendicular to the plane of the loop.

Opposite to the current.

In the plane of the loop.

Correct Answer: B

Solution:

The direction of the magnetic moment

m

is perpendicular to the plane of the loop, given by the right-hand thumb rule.

4.0 \times 10^{-6} \text{ T}

2.0 \times 10^{-6} \text{ T}

8.0 \times 10^{-6} \text{ T}

1.0 \times 10^{-6} \text{ T}

Correct Answer: B

Solution:

Using the formula

B = \frac{\mu_0 I}{2\pi R}

, where

I = 10 \text{ A}

and

R = 0.05 \text{ m}

, we get

B = \frac{4\pi \times 10^{-7} \times 10}{2\pi \times 0.05} = 2.0 \times 10^{-6} \text{ T}

They attract each other.

They repel each other.

They have no effect on each other.

They cancel each other out.

Correct Answer: A

Solution:

Parallel currents attract each other due to the magnetic fields they generate.

Strong and uniform

Weak and uniform

Strong and non-uniform

Weak and non-uniform

Correct Answer: D

Solution:

Outside a long solenoid, the magnetic field is weak and non-uniform, as the field lines spread out.

Parallel to the axis of the solenoid

Perpendicular to the axis of the solenoid

Circular around the solenoid

Randomly oriented

Correct Answer: A

Solution:

Inside a long solenoid, the magnetic field is uniform and parallel to the axis of the solenoid.

0.002 T

0.004 T

0.006 T

0.008 T

Correct Answer: B

Solution:

The magnetic field inside a long solenoid is given by

B = \mu_0 n I

. Substituting the given values:

\mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}

n = 1000 \text{ turns/m}

, and

I = 2 \text{ A}

, we get

B = 0.004 \text{ T}

The Lorentz force is independent of the velocity of the charged particle.

The Lorentz force depends only on the magnetic field and not on the electric field.

The Lorentz force can be zero if the velocity of the charged particle is parallel to the magnetic field.

The Lorentz force is always perpendicular to the velocity of the charged particle.

Correct Answer: C

Solution:

The Lorentz force is given by

F = q (v \times B + E)

. If the velocity

v

is parallel to the magnetic field

B

, the cross product

v \times B

is zero, making the magnetic part of the force zero.

4.0 × 10⁻⁶ T

2.0 × 10⁻⁶ T

1.0 × 10⁻⁶ T

5.0 × 10⁻⁶ T

Correct Answer: B

Solution:

Using Ampere's Circuital Law, the magnetic field

B

at a distance

R

from a long straight wire carrying current

I

is given by:

B = \frac{\mu_0 I}{2\pi R}

where

\mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A}

. Substituting the values,

B = \frac{4\pi \times 10^{-7} \times 50}{2\pi \times 2.5} = 2.0 \times 10^{-6} \, \text{T}

B = 5 \times 10^{-4} \text{ T}

B = 2 \times 10^{-4} \text{ T}

B = 1 \times 10^{-4} \text{ T}

B = 4 \times 10^{-4} \text{ T}

Correct Answer: A

Solution:

The magnetic field at the center of a circular coil is given by

B = \frac{\mu_0 N I}{2R}

, where

N

is the number of turns,

I

is the current, and

R

is the radius. Substituting the given values,

B = \frac{4\pi \times 10^{-7} \times 200 \times 0.2}{2 \times 0.05} = 5 \times 10^{-4} \text{ T}

It is independent of the area of the loop.

It is proportional to the number of turns

N

and the area

A

of the loop.

It is inversely proportional to the current

I

It is zero if the loop is placed in a uniform magnetic field.

Correct Answer: B

Solution:

The magnetic moment

m

of a planar loop is given by

m = NIA

, where

N

is the number of turns and

A

is the area of the loop.

0.157 N·m

0.314 N·m

0.628 N·m

0 N·m

Correct Answer: D

Solution:

The torque on a current-carrying loop in a magnetic field is given by

\tau = m \times B

, where

m = I \cdot A

is the magnetic moment and

A

is the area of the loop. Since the plane of the loop is perpendicular to the magnetic field, the angle between

m

and

B

90^\circ

, making the torque

\tau = 0

It increases

It decreases

It remains constant

It becomes zero

Correct Answer: B

Solution:

The magnetic field strength decreases as the distance from a long straight wire carrying current increases.

They are straight lines parallel to the wire

They form concentric circles around the wire

They radiate outward from the wire

They do not exist around the wire

Correct Answer: B

Solution:

The magnetic field lines around a long, straight current-carrying wire form concentric circles with the wire at the center.

9.58 \times 10^5 m/s

1.92 \times 10^6 m/s

3.84 \times 10^6 m/s

4.79 \times 10^5 m/s

Correct Answer: A

Solution:

The centripetal force required for circular motion is provided by the magnetic force:

qvB = \frac{mv^2}{r}

. Solving for

v

, we have

v = \frac{qBr}{m}

. Substituting

q = 1.6 \times 10^{-19} C

B = 0.5 T

r = 0.1 m

, and

m = 1.67 \times 10^{-27} kg

, we get

v = \frac{1.6 \times 10^{-19} \times 0.5 \times 0.1}{1.67 \times 10^{-27}} = 9.58 \times 10^5 m/s

They are straight lines parallel to the wire.

They form concentric circles around the wire.

They radiate outwards from the wire.

They form elliptical paths around the wire.

Correct Answer: B

Solution:

The magnetic field lines produced by a straight current-carrying wire form concentric circles around the wire, as observed in Oersted's experiment.

Faraday's Law

Ampere's Circuital Law

Biot-Savart Law

Gauss's Law

Correct Answer: B

Solution:

Ampere's Circuital Law states that the line integral of the magnetic field around a closed loop is proportional to the current passing through the loop.

They repel each other.

They attract each other.

There is no force between them.

The force depends on the distance between the wires.

Correct Answer: B

Solution:

Parallel currents attract each other due to the magnetic fields generated by the currents. This is a consequence of Ampere's force law.

Magnetic field lines form closed loops.

Magnetic field lines originate from positive charges.

Magnetic field lines terminate at negative charges.

Magnetic field lines are always straight.

Correct Answer: A

Solution:

Magnetic field lines form closed loops, unlike electric field lines which originate from positive charges and terminate at negative charges.

B = \frac{\mu_0 I}{2\pi R}

B = \frac{\mu_0 I}{4\pi R}

B = \frac{\mu_0 I}{\pi R}

B = \frac{\mu_0 I}{R}

Correct Answer: A

Solution:

According to Ampere's Circuital Law, the magnetic field due to a long straight wire is given by

B = \frac{\mu_0 I}{2\pi R}

, where

\mu_0

is the permeability of free space,

I

is the current, and

R

is the distance from the wire.

2 \times 10^{-5} N/m, attractive

2 \times 10^{-5} N/m, repulsive

4 \times 10^{-5} N/m, attractive

4 \times 10^{-5} N/m, repulsive

Correct Answer: B

Solution:

The force per unit length between two parallel currents is given by

F/L = \frac{\mu_0 I_1 I_2}{2\pi d}

. Substituting

I_1 = I_2 = 10 \text{ A}

and

d = 0.1 \text{ m}

, we get

F/L = \frac{4\pi \times 10^{-7} \times 10 \times 10}{2\pi \times 0.1} = 2 \times 10^{-5} \text{ N/m}

. Since the currents are in opposite directions, the force is repulsive.

\tau = NIAB

\tau = 0

\tau = \frac{NIAB}{2}

\tau = 2NIAB

Correct Answer: B

Solution:

The torque on a current-carrying loop in a magnetic field is given by

\tau = m \times B

, where

m = NI\vec{A}

. When the loop is parallel to the magnetic field, the angle between

m

and

B

is 0, making the torque zero.

1 \times 10^{-5} T

3 \times 10^{-5} T

2 \times 10^{-5} T

5 \times 10^{-5} T

Correct Answer: C

Solution:

The magnetic field due to a long straight wire is given by

B = \frac{\mu_0 I}{2\pi R}

. Substituting

I = 15 A

R = 0.3 m

, and

\mu_0 = 4\pi \times 10^{-7} Tm/A

, we get

B = \frac{4\pi \times 10^{-7} \times 15}{2\pi \times 0.3} = 2 \times 10^{-5} T

\tau = 0

\tau = N I A B

\tau = N I B R

\tau = N I B R^2

Correct Answer: A

Solution:

When the plane of the loop is parallel to the magnetic field, the angle between the magnetic moment and the field is zero, resulting in zero torque (

\tau = m \times B = 0

Right-hand rule

Left-hand rule

Fleming's rule

Newton's rule

Correct Answer: A

Solution:

The right-hand rule is used to determine the direction of the magnetic field around a current-carrying wire by pointing the thumb in the direction of the current and curling the fingers around the wire.

1.26 \times 10^{-3} \text{ T}

1.26 \times 10^{-2} \text{ T}

1.26 \times 10^{-4} \text{ T}

1.26 \times 10^{-5} \text{ T}

Correct Answer: A

Solution:

Using the formula

B = \mu_0 n I

, where

n = 500 \text{ turns/m}

and

I = 2 \text{ A}

, we get

B = 4\pi \times 10^{-7} \times 500 \times 2 = 1.26 \times 10^{-3} \text{ T}

It remains the same.

It doubles.

It halves.

It becomes zero.

Correct Answer: C

Solution:

The magnetic field inside a solenoid is given by

B = \mu_0 n I

, where

n

is the number of turns per unit length. If the solenoid is stretched to twice its original length,

n

becomes half. Therefore, the magnetic field also halves.

The magnetic moment decreases

The magnetic moment remains the same

The magnetic moment increases

The magnetic moment becomes zero

Correct Answer: C

Solution:

The magnetic moment

m

of a planar loop is given by

m = NIA

, where

N

is the number of turns. Increasing

N

increases the magnetic moment.

B.dl = \mu_0 I

B = \frac{\mu_0 I}{2\pi R}

B = \mu_0 n I

B = \frac{1}{2R}

Correct Answer: A

Solution:

Ampere's Circuital Law states that the line integral of the magnetic field

B

around a closed loop is equal to

\mu_0

times the current

I

passing through the loop.

k\Phi = NIAB

k\Phi = \frac{NIB}{A}

k\Phi = \frac{NAB}{I}

k\Phi = \frac{IAB}{N}

Correct Answer: A

Solution:

The equilibrium deflection

\Phi

in a moving coil galvanometer is given by the equation

k\Phi = NIAB

, where

k

is the torsion constant,

N

is the number of turns,

I

is the current,

A

is the area, and

B

is the magnetic field.

By introducing a shunt resistance of small value in parallel.

By introducing a large resistance in series.

By removing the coil.

By increasing the coil's turns.

Correct Answer: A

Solution:

A moving coil galvanometer can be converted into an ammeter by introducing a shunt resistance of small value in parallel.

The field lines disappear.

The field lines remain unchanged.

The field lines reverse direction.

The field lines become concentric ellipses.

Correct Answer: C

Solution:

When the current is reversed, the direction of the magnetic field lines also reverses, as observed by Oersted.

B = \frac{\mu_0 I}{2\pi R}

BL = \mu_0 I_e

B = \mu_0 n I

B = 0

Correct Answer: B

Solution:

Ampere's Circuital Law states that the line integral of the magnetic field

B

around a closed loop is equal to the permeability of free space

\mu_0

times the current

I_e

enclosed by the loop.

m = NIA

m = \frac{N}{A}

m = \frac{A}{NI}

m = NI^2A

Correct Answer: A

Solution:

The magnetic moment

m

of a planar loop is given by

m = NIA

, where

N

is the number of turns,

I

is the current, and

A

is the area of the loop.

0.0945 \text{ A m}^2

0.0945 \text{ A m}

0.0945 \text{ A m}^3

0.0945 \text{ A m}^{-1}

Correct Answer: A

Solution:

The area

A

of the loop is

\pi r^2 = \pi \times (0.1)^2 = 0.0314 \text{ m}^2

. The magnetic moment

m = IA = 3 \times 0.0314 = 0.0945 \text{ A m}^2

The magnetic field remains the same

The magnetic field is halved

The magnetic field doubles

The magnetic field becomes zero

Correct Answer: C

Solution:

The magnetic field inside a long solenoid is directly proportional to the current, so doubling the current doubles the magnetic field.

It remains the same

It doubles

It halves

It becomes zero

Correct Answer: B

Solution:

The magnetic field at the center of a circular coil is directly proportional to the current. Thus, doubling the current doubles the magnetic field.

m = I/A

m = I \times A

m = A/I

m = I^2 \times A

Correct Answer: B

Solution:

The magnetic moment

m

of a planar loop carrying current

I

is given by

m = I \times A

, where

A

is the area of the loop.

By introducing a large resistance in series.

By introducing a small resistance in parallel.

By increasing the current through the coil.

By decreasing the number of turns in the coil.

Correct Answer: A

Solution:

A moving coil galvanometer can be converted into a voltmeter by introducing a large resistance in series with the coil.

1.0 \times 10^{-5} \text{ N/m}

2.0 \times 10^{-5} \text{ N/m}

3.0 \times 10^{-5} \text{ N/m}

4.0 \times 10^{-5} \text{ N/m}

Correct Answer: B

Solution:

Using

F/L = \frac{\mu_0 I_1 I_2}{2\pi d}

, where

I_1 = 5 \text{ A}

I_2 = 10 \text{ A}

, and

d = 0.1 \text{ m}

, we get

F/L = \frac{4\pi \times 10^{-7} \times 5 \times 10}{2\pi \times 0.1} = 2.0 \times 10^{-5} \text{ N/m}

0.02 T

0.04 T

0.06 T

0.08 T

Correct Answer: A

Solution:

Using Ampere's Circuital Law, the magnetic field at a distance

R

from a long straight wire is given by

B = \frac{\mu_0 I}{2\pi R}

. Substituting the given values:

\mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}

I = 10 \text{ A}

, and

R = 0.1 \text{ m}

, we get

B = 0.02 \text{ T}

1.6 x 10^-5 T

2.0 x 10^-5 T

3.2 x 10^-5 T

4.0 x 10^-5 T

Correct Answer: A

Solution:

The magnetic field at a distance

R

from a long straight wire carrying a current

I

is given by

B = \frac{\mu_0 I}{2 \pi R}

. Substituting

\mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}

I = 40 \text{ A}

, and

R = 0.5 \text{ m}

, we get

B = \frac{4\pi \times 10^{-7} \times 40}{2\pi \times 0.5} = 1.6 \times 10^{-5} \text{ T}

It is strongest outside the solenoid.

It is strongest at the ends of the solenoid.

It is uniform and strong inside the solenoid.

It is zero inside the solenoid.

Correct Answer: C

Solution:

The magnetic field inside a long solenoid is uniform and strong, as the field lines are parallel and closely spaced.

The magnetic field lines inside the solenoid are parallel and equally spaced.

The magnetic field lines inside the solenoid form concentric circles.

The magnetic field lines inside the solenoid are radial.

The magnetic field lines inside the solenoid are perpendicular to the axis of the solenoid.

Correct Answer: A

Solution:

Inside a long solenoid, the magnetic field lines are parallel and equally spaced, indicating a uniform magnetic field along the axis of the solenoid.

5 x 10^-6 N/m

1 x 10^-5 N/m

2.5 x 10^-5 N/m

3 x 10^-5 N/m

Correct Answer: B

Solution:

The force per unit length between two parallel wires is given by

F/L = \frac{\mu_0 I_1 I_2}{2\pi d}

, where

I_1

and

I_2

are the currents in the wires, and

d

is the distance between them. Substituting

\mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}

I_1 = I_2 = 5 \text{ A}

, and

d = 0.2 \text{ m}

, we get

F/L = \frac{4\pi \times 10^{-7} \times 5 \times 5}{2\pi \times 0.2} = 1 \times 10^{-5} \text{ N/m}

1.0 \times 10^{-4} \text{ T}

2.0 \times 10^{-4} \text{ T}

3.0 \times 10^{-4} \text{ T}

4.0 \times 10^{-4} \text{ T}

Correct Answer: A

Solution:

The magnetic field at the center of a circular coil is given by

B = \frac{\mu_0 N I}{2R}

. Substituting

N = 100

I = 0.40

R = 0.08

m, and

\mu_0 = 4\pi \times 10^{-7}

Tm/A, we get

B = 1.0 \times 10^{-4}

1.26 x 10^-3 T

1.57 x 10^-3 T

2.51 x 10^-3 T

3.14 x 10^-3 T

Correct Answer: A

Solution:

The magnetic field inside a solenoid is given by

B = \mu_0 n I

, where

n

is the number of turns per unit length and

I

is the current. Substituting

\mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}

n = 500 \text{ m}^{-1}

, and

I = 2 \text{ A}

, we get

B = 4\pi \times 10^{-7} \times 500 \times 2 = 1.26 \times 10^{-3} \text{ T}

Radial from the wire

Concentric circles around the wire

Parallel to the wire

Perpendicular to the wire

Correct Answer: B

Solution:

The magnetic field lines around a long straight wire carrying a current form concentric circles with the wire at the center.

To increase the sensitivity of the galvanometer

To decrease the sensitivity of the galvanometer

To allow the galvanometer to measure larger currents

To protect the galvanometer from high voltages

Correct Answer: C

Solution:

A shunt resistance of small value is introduced in parallel with the galvanometer to allow most of the current to bypass the galvanometer coil, enabling it to measure larger currents without being damaged.

The magnetic field lines are straight and parallel to the wire.

The magnetic field lines form concentric circles around the wire.

The magnetic field lines are perpendicular to the wire and point radially outward.

The magnetic field lines form a spiral pattern around the wire.

Correct Answer: B

Solution:

The magnetic field lines around a current-carrying wire form concentric circles, as described by the right-hand rule.

It remains the same.

It doubles.

It halves.

It quadruples.

Correct Answer: B

Solution:

The torque in a moving coil galvanometer is given by

\tau = NIAB

, where

N

is the number of turns. If

N

is doubled, the torque doubles, leading to a doubling of the equilibrium deflection angle

\Phi

, assuming the spring constant

k

remains unchanged.

I = \frac{k\Phi}{NAB}

I = \frac{NAB}{k\Phi}

I = \frac{NABk}{\Phi}

I = \frac{\Phi}{NABk}

Correct Answer: A

Solution:

The torque

\tau

in a moving coil galvanometer is given by

\tau = NIAB

. This is balanced by the counter-torque from the spring,

k\Phi

. Therefore,

NIAB = k\Phi

. Solving for

I

, we get

I = \frac{k\Phi}{NAB}

3 A⋅m²

1.5 A⋅m²

2.0 A⋅m²

1.0 A⋅m²

Correct Answer: A

Solution:

The magnetic moment

m

of a loop is given by

m = NIA

, where

N

is the number of turns,

I

is the current, and

A

is the area. Substituting

N = 20

I = 3 \text{ A}

, and

A = 0.05 \text{ m}^2

, we get

m = 20 \times 3 \times 0.05 = 3 \text{ A⋅m}^2

k\Phi = NIAB

k\Phi = \frac{NI}{AB}

k\Phi = \frac{NAB}{I}

k\Phi = \frac{I}{NAB}

Correct Answer: A

Solution:

In a moving coil galvanometer, the equilibrium condition is given by

k\Phi = NIAB

, where

k

is the torsion constant,

\Phi

is the deflection,

N

is the number of turns,

I

is the current,

A

is the area, and

B

is the magnetic field.

9.42 x 10^-4 T

1.57 x 10^-3 T

2.36 x 10^-3 T

3.14 x 10^-3 T

Correct Answer: A

Solution:

The magnetic field at the center of a circular coil is given by

B = \frac{\mu_0 N I}{2R}

, where

N

is the number of turns,

I

is the current, and

R

is the radius. Substituting

\mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}

N = 150

I = 0.5 \text{ A}

, and

R = 0.1 \text{ m}

, we get

B = \frac{4\pi \times 10^{-7} \times 150 \times 0.5}{2 \times 0.1} = 9.42 \times 10^{-4} \text{ T}

\tau = k\Phi

\tau = NIAB

\tau = \frac{NIAB}{k}

\tau = \frac{k\Phi}{NIAB}

Correct Answer: B

Solution:

In a moving coil galvanometer, the magnetic torque is given by

\tau = NIAB

, where

N

is the number of turns,

I

is the current,

A

is the area, and

B

is the magnetic field. This torque is balanced by the counter-torque

k\Phi

, leading to the equilibrium condition

k\Phi = NIAB

They form concentric circles around the wire.

They radiate outwards in straight lines.

They form a spiral pattern along the wire.

They are parallel to the wire.

Correct Answer: A

Solution:

The magnetic field lines around a long, straight current-carrying wire form concentric circles with the wire at the center.

F/L = B I

F/L = 2 \pi B I

F/L = B I \sin \theta

F/L = 0

Correct Answer: A

Solution:

The force per unit length on a current-carrying wire in a magnetic field is given by

F/L = B I \sin \theta

. Since the field is perpendicular,

\theta = 90^\circ

, so

\sin \theta = 1

. Thus,

F/L = B I

It is directly proportional to

R

It is inversely proportional to

R

It is independent of

R

It is proportional to

R^2

Correct Answer: B

Solution:

The magnetic field at a distance

R

from a long, straight wire is inversely proportional to

R

, as given by the formula

B = \frac{\mu_0 I}{2\pi R}

It remains the same

It doubles

It halves

It becomes zero

Correct Answer: C

Solution:

The magnetic field inside a solenoid is given by

B = \mu_0 n I

, where

n

is the number of turns per unit length. When the solenoid is stretched to twice its length, the number of turns per unit length

n

halves, thus halving the magnetic field.

Clockwise when viewed from above

Counter-clockwise when viewed from above

Radially inwards

Radially outwards

Correct Answer: B

Solution:

According to the right-hand rule, if you point your thumb in the direction of the current (upwards), your fingers curl in the direction of the magnetic field, which is counter-clockwise when viewed from above.

The magnetic field doubles.

The magnetic field halves.

The magnetic field remains the same.

The magnetic field becomes zero.

Correct Answer: A

Solution:

The magnetic field inside a solenoid is given by

B = \mu_0 n I

, where

n

is the number of turns per unit length. Doubling

n

while keeping

I

constant results in the magnetic field doubling.

B = \frac{\mu_0 I r}{2\pi a^2}

B = \frac{\mu_0 I}{2\pi r}

B = \frac{\mu_0 I r^2}{2\pi a^2}

B = \frac{\mu_0 I a}{2\pi r^2}

Correct Answer: A

Solution:

Using Ampere's Law, for

r < a

, the magnetic field

B

is given by

B(2\pi r) = \mu_0 I_e

, where

I_e = I \left(\frac{\pi r^2}{\pi a^2}\right) = \frac{Ir^2}{a^2}

. Thus,

B = \frac{\mu_0 I r}{2\pi a^2}

The field lines are straight and parallel to the wire

The field lines form concentric circles around the wire

The field lines are random and disorganized

The field lines are perpendicular to the wire

Correct Answer: B

Solution:

The magnetic field lines around a straight current-carrying wire form concentric circles centered on the wire.

True or False

Correct Answer: True

Solution:

Magnetic field lines always form closed loops, whereas electrostatic field lines originate from positive charges and terminate at negative charges.

Correct Answer: False

Solution:

The magnetic field produced by a straight current-carrying wire is inversely proportional to the distance from the wire, not directly proportional.

Correct Answer: True

Solution:

The Lorentz force is given by the expression

F = q (v \times B + E)

, where it depends on the velocity

v

of the charged particle.

Correct Answer: False

Solution:

The magnetic field inside a long solenoid is uniform and directed along the axis of the solenoid, with no perpendicular component.

Correct Answer: False

Solution:

The Lorentz force is perpendicular to both the velocity of the charged particle and the magnetic field, not along the velocity.

Correct Answer: False

Solution:

Parallel currents attract each other, while anti-parallel currents repel.

Correct Answer: True

Solution:

Correct Answer: False

Solution:

The magnetic field due to a long straight wire is not infinite at any non-zero distance; it is directly proportional to the current and inversely proportional to the distance from the wire.

Correct Answer: False

Solution:

A moving coil galvanometer is converted into a voltmeter by introducing a large resistance in series, not a small shunt resistance in parallel.

Correct Answer: True

Solution:

The force on a planar loop carrying a current in a uniform magnetic field is zero, as mentioned in the excerpts.

Correct Answer: False

Solution:

Ampere's Circuital Law is not independent of the Biot-Savart Law; it can be derived from it, similar to how Gauss's Law relates to Coulomb's Law.

Correct Answer: True

Solution:

In a moving coil galvanometer, the torque due to the magnetic field, given by

\tau = m \times B

, is balanced by a counter-torque from a spring, resulting in an equilibrium deflection.

Correct Answer: True

Solution:

The net force on a planar loop carrying a current in a uniform magnetic field is zero, but it can experience a torque.

Correct Answer: True

Solution:

Inside a long solenoid, the magnetic field is uniform, strong, and directed along the axis of the solenoid.

Correct Answer: True

Solution:

Magnetic field lines form closed loops, unlike electrostatic field lines which start and end at charges.

Correct Answer: False

Solution:

The Lorentz force is dependent on the velocity of the particle, as it is given by the equation

F = q (v \times B + E)

, where

v

is the velocity.

Correct Answer: False

Solution:

Parallel currents attract each other, while anti-parallel currents repel.

Correct Answer: True

Solution:

The magnetic field inside a long solenoid is given by

B = \mu_0 n I

, where

n

is the number of turns per unit length, and

I

is the current. This results in a uniform magnetic field inside the solenoid.

Correct Answer: False

Solution:

The Lorentz force depends on the velocity of the charged particle, as it is given by the expression

F = q (v \times B + E)

, where

v

is the velocity.

Correct Answer: True

Solution:

A current-carrying loop in a uniform magnetic field experiences no net force, but it can experience a torque due to the magnetic moment interacting with the field.

Correct Answer: True

Solution:

A moving coil galvanometer can be converted into an ammeter by introducing a shunt resistance of small value in parallel, as stated in the excerpts.

Correct Answer: False

Solution:

A moving coil galvanometer is converted into a voltmeter by introducing a large resistance in series, not a shunt resistance in parallel.

Correct Answer: True

Solution:

The magnetic field inside a long solenoid is uniform and directed along the axis, making it strong and consistent.

Correct Answer: False

Solution:

Ampere's Circuital Law is not independent of the Biot-Savart Law; it can be derived from it, similar to how Gauss's Law relates to Coulomb's Law.

Correct Answer: True

Solution:

The magnetic moment

m

of a planar loop with

N

turns and area

A

is given by

m = NA

. Its direction is determined by the right-hand thumb rule.

Correct Answer: True

Solution:

The magnetic field lines around a long straight wire carrying current are concentric circles with the wire at the center, as depicted in Oersted's experiments and described in the excerpts.

Correct Answer: True

Solution:

Ampere's Circuital Law is not independent of the Biot-Savart law and can be derived from it, similar to how Gauss's law relates to Coulomb's law.

Correct Answer: False

Solution:

The permeability of free space, denoted as

\mu_0

, is a scalar quantity, not a vector.

Correct Answer: False

Solution:

The magnetic field inside a long solenoid is strong and uniform along its axis, as described in the excerpts.

Correct Answer: True

Solution:

Magnetic field lines always form closed loops, as they do not have a beginning or an end, unlike electrostatic field lines which start from positive charges and end at negative charges.

Correct Answer: True

Solution:

The right-hand rule is a method to determine the direction of the magnetic field around a straight current-carrying wire, as mentioned in the excerpts.

Correct Answer: False

Solution:

Parallel currents attract each other, while anti-parallel currents repel.

Correct Answer: True

Solution:

Inside a long solenoid, the magnetic field lines are uniform and parallel to the axis of the solenoid.

Correct Answer: False

Solution:

Parallel currents attract each other while anti-parallel currents repel.

Correct Answer: True

Solution:

The right-hand rule is used to determine the direction of the magnetic field around a current-carrying wire: by pointing the thumb in the direction of the current, the fingers curl in the direction of the magnetic field.

Correct Answer: True

Solution:

Hans Christian Oersted observed that a current in a straight wire causes a noticeable deflection in a nearby magnetic compass needle, demonstrating the relationship between electricity and magnetism.

Correct Answer: True

Solution:

Magnetic field lines form closed loops, whereas electrostatic field lines originate from positive charges and terminate at negative charges or fade at infinity.

Correct Answer: False

Solution:

Ampere's Circuital Law relates the magnetic field around a closed loop to the current passing through the loop, not the electric field.

Correct Answer: True

Solution:

The direction of the magnetic moment in a planar loop is given by the right-hand thumb rule, where the thumb points in the direction of the magnetic moment when the fingers curl in the direction of the current.

Correct Answer: True

Solution:

Ampere's Circuital Law is expressed as

\oint B \cdot dl = \mu_0 I

, where the integral of the magnetic field

B

around a closed loop is equal to the permeability of free space

\mu_0

times the current

I

enclosed by the loop.

Correct Answer: False

Solution:

The magnetic field due to a long straight wire is not infinite at a non-zero distance; it is proportional to the current and inversely proportional to the distance from the wire.

Correct Answer: False

Solution:

Magnetic field lines form closed loops around a current-carrying wire, as opposed to electrostatic field lines which start and end on charges.

Correct Answer: True

Solution:

Oersted's experiment demonstrated the relationship between electricity and magnetism by showing that a current can produce a magnetic field.

Correct Answer: True

Solution:

The magnetic field lines around a long straight current-carrying wire are indeed concentric circles centered on the wire, as described in the excerpts.

Correct Answer: False

Solution:

The direction of the magnetic moment of a planar loop carrying current is given by the right-hand thumb rule, not the left-hand rule.

Correct Answer: False

Solution:

The magnetic field inside a long solenoid is actually strong and uniform along the axis of the solenoid, contrary to the statement.

Correct Answer: True

Solution:

Ampere's Circuital Law can indeed be derived from the Biot-Savart Law, as indicated in the excerpts.

Correct Answer: True

Solution:

The magnetic moment of a planar loop carrying a current is indeed given by the product of the number of turns and the area of the loop, as stated in the excerpts.

Correct Answer: False

Solution:

Ampere's Circuital Law is valid only for steady currents. When currents vary with time, the momentum carried by the electromagnetic field must be considered.

Correct Answer: True

Solution:

The magnetic field inside a long solenoid is uniform and strong along its axis, while outside the solenoid, the field is weak and non-uniform, as described in the provided excerpts.

Moving Charges and Magnetism

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Summary

Chapter Summary: Moving Charges and Magnetism

Key Concepts

Important Formulas

Learning Objectives

Common Mistakes & Exam Tips

Important Diagrams

Learning Objectives

Learning Objectives

Detailed Notes

Chapter Notes: Moving Charges and Magnetism

1. Introduction

2. Physical Quantities

3. Key Concepts

3.1 Electrostatic and Magnetic Fields

3.2 Lorentz Force

3.3 Ampere's Circuital Law

3.4 Magnetic Field from a Long Straight Wire

3.5 Magnetic Field Inside a Solenoid

4. Important Relationships

5. Exercises

6. Points to Ponder

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Tips for Success

Practice & Assessment

Multiple Choice Questions

Q1.What did Hans Christian Oersted discover about the relationship between electricity and magnetism?

Correct Answer: A

Q2.What is the effect of parallel currents on each other?

Correct Answer: B

Q3.What is the effect of introducing a shunt resistance in parallel with a moving coil galvanometer?

Correct Answer: C

Q4.In a moving coil galvanometer, what is the role of the torsion constant kkk?

Correct Answer: B

Q5.What happens to the magnetic field inside a long solenoid when the number of turns per unit length increases?

Correct Answer: A

Q6.A long solenoid has nnn turns per unit length and carries a current III. What is the expression for the magnetic field inside the solenoid?

Correct Answer: A

Q7.A long straight wire carries a current of 50 A in the north to south direction. Using the right-hand rule, what is the direction of the magnetic field at a point 2.5 m east of the wire?

Correct Answer: A

Q8.Which of the following statements is true about the magnetic field inside a long solenoid?

Correct Answer: B

Q9.What is the direction of the magnetic moment mmm of a planar loop carrying current III?

Correct Answer: B

Correct Answer: B

Q11.What happens to parallel currents according to the principles of magnetism?

Correct Answer: A

Q12.Which of the following best describes the magnetic field outside a long solenoid?

Correct Answer: D

Q13.What is the direction of the magnetic field inside a long solenoid?

Correct Answer: A

Q14.A long solenoid has 1000 turns per meter and carries a current of 2 A. What is the magnitude of the magnetic field inside the solenoid?

Correct Answer: B

Q15.Which of the following statements about the Lorentz force is true?

Correct Answer: C

Q16.A long straight wire carries a current of 50 A. Using Ampere's Circuital Law, what is the magnitude of the magnetic field at a point 2.5 m away from the wire?

Correct Answer: B

Q17.A circular coil with 200 turns and a radius of 5 cm carries a current of 0.2 A. What is the magnitude of the magnetic field at the center of the coil?

Correct Answer: A

Q18.Which of the following statements is true about the magnetic moment mmm of a planar loop carrying current III?

Correct Answer: B

Q19.A circular loop of radius 0.5 m carries a current of 2 A. It is placed in a uniform magnetic field of 0.1 T such that the plane of the loop is perpendicular to the magnetic field. What is the torque acting on the loop?

Correct Answer: D

Q20.What happens to the magnetic field strength at a point when the distance from a long straight wire carrying current increases?

Correct Answer: B

Q21.Which of the following statements is true about the magnetic field lines around a long, straight current-carrying wire?

Correct Answer: B

Q22.A proton moves in a circular path of radius 0.1 m in a magnetic field of 0.5 T. If the velocity of the proton is perpendicular to the magnetic field, what is the speed of the proton?

Correct Answer: A

Q23.Which of the following statements is true about the magnetic field lines produced by a straight current-carrying wire?

Correct Answer: B

Q24.Which law relates the magnetic field around a closed loop to the current passing through the loop?

Correct Answer: B

Q25.Two parallel wires carry currents I1I_1I1​ and I2I_2I2​ in the same direction. What is the nature of the force between them?

Correct Answer: B

Q26.Which of the following statements about the magnetic field lines is true?