Moving Charges and Magnetism

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Learning Objectives

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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.

Revision Notes & Summary

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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

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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.

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Practice Test – MCQs, True/False

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Multiple Choice Questions

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}

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.

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}

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

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.

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.

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.

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

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.

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.

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.

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.

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.

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

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.

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.

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.

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}

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.

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}

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 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.

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}

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.

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}

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.

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.

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.

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.

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 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.

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.

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

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.

\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

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

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}

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.

\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.

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 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.

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.

\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

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}

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.

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

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.

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.

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.

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.

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}

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.

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.

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}

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.

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.

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.

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.

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}

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.

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.

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.

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.

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.

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.

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.

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}

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.

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.

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:

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:

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

Solution:

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

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: 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 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:

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:

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: 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:

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: 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:

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

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: 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:

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:

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

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.

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.

Moving Charges and Magnetism

Learning Objectives

Learning Objectives

Revision Notes & Summary

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

AI Learning Assistant

Practice Test – MCQs, True/False

Multiple Choice Questions

Q1.Two parallel wires, each carrying a current of 5 A, are placed 0.2 m apart. What is the force per unit length between the wires?

Correct Answer: B

Q2.In a moving coil galvanometer, the torque due to the magnetic field is balanced by a counter-torque from a spring. Which expression represents this equilibrium?

Correct Answer: A

Q3.In a hypothetical scenario, a long straight wire carries a current of 40 A. A point is located at a distance of 0.5 m from the wire. Calculate the magnitude of the magnetic field at this point using the formula for the magnetic field around a long straight wire.

Correct Answer: A

Q4.A long straight wire carries a current of 15 A. What is the magnetic field at a distance of 0.3 m from the wire?

Correct Answer: C

Q5.What happens to the magnetic field strength at a point when the distance from a long straight wire carrying current increases?

Correct Answer: B

Q6.What happens to the magnetic field lines around a long, straight wire when the current is reversed?

Correct Answer: C

Q7.Which of the following statements is true about the magnetic field inside a long solenoid?

Correct Answer: B

Q8.Consider a long, straight wire carrying a current III. If the wire is placed in a uniform magnetic field BBB perpendicular to the current, what is the force per unit length on the wire?

Correct Answer: A

Q9.How can a moving coil galvanometer be converted into a voltmeter?

Correct Answer: A

Q10.Which of the following statements is true about the magnetic moment mmm of a planar loop carrying current III?

Correct Answer: B

Q11.A solenoid is stretched to twice its original length while maintaining the same number of turns and current. What happens to the magnetic field inside the solenoid?

Correct Answer: C

Q12.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

Q13.Which rule can be used to determine the direction of the magnetic field due to a current-carrying wire?

Correct Answer: A

Q14.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

Q15.A moving coil galvanometer is converted into an ammeter by introducing a shunt resistance. What is the purpose of this shunt resistance?

Correct Answer: C

Q16.How can a moving coil galvanometer be converted into an ammeter?

Correct Answer: A

Q17.Consider a long solenoid with 1000 turns per meter carrying a current of 5 A. If the solenoid is stretched to twice its original length without changing the number of turns or the current, what will be the new magnetic field inside the solenoid?

Correct Answer: C

Q18.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

Q19.What is the direction of the magnetic field inside a long solenoid?

Correct Answer: A

Q20.A circular coil with 150 turns and a radius of 10 cm carries a current of 0.5 A. Calculate the magnetic field at the center of the coil.

Correct Answer: A

Q21.How does the magnetic field at a distance RRR from a long, straight wire carrying a current III vary with RRR?

Correct Answer: B

Q22.In a moving coil galvanometer, the torque due to the magnetic field is balanced by a counter-torque from a spring. If the number of turns in the coil is doubled, what happens to the equilibrium deflection angle Φ\PhiΦ?

Correct Answer: B

Q23.A solenoid with 500 turns per meter carries a current of 2 A. What is the magnitude of the magnetic field inside the solenoid? Use the formula B=μ0nIB = \mu_0 n IB=μ0​nI.

Correct Answer: A

Q24.According to Ampere's Circuital Law, what is the relationship between the magnetic field BBB and the current III enclosed by a loop?

Correct Answer: B

Q25.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

Q26.What happens to the magnetic field inside a long solenoid if the current is doubled?

Correct Answer: C

Q27.Consider a long straight wire carrying a current III. According to Ampere's Circuital Law, what is the expression for the magnetic field BBB at a distance RRR from the wire?

Correct Answer: A

Q28.What is the effect of increasing the number of turns NNN in a planar loop carrying current III on its magnetic moment mmm?

Correct Answer: C

Q29.What happens to the magnetic field at the center of a circular coil when the current is doubled?