NEET Physics: Magnetism, Electromagnetic Induction and AC · Chapter 1

Moving Charges and Magnetism

What to remember

A magnetic force acts only on a moving charge: F = q v × B, magnitude F = qvB sinθ. It is always perpendicular to both v and B, so it does no work and never changes the speed.
In a uniform field a charge moving perpendicular to B goes in a circle of radius r = mv/qB with period T = 2πm/qB. The period does not depend on speed or radius.
Ampere's circuital law gives B = μ₀I/2πr for a long straight wire and B = μ₀nI inside a long solenoid. Two parallel wires carrying currents in the same direction attract each other.
A galvanometer becomes an ammeter with a small shunt in parallel and a voltmeter with a large resistance in series. Torque on a loop is τ = NIAB sinθ = m B sinθ.

Oersted's experiment and the magnetic force on a moving charge

Oersted found that a compass needle placed near a wire is deflected when the wire carries a current. The needle turns the other way when the current is reversed. This showed that an electric current produces a magnetic field around it.

A charge q moving with velocity v in a magnetic field B experiences the magnetic force

F = q (v × B), with magnitude F = q v B sinθ

where θ is the angle between v and B. The unit of B is the tesla (T = N A⁻¹ m⁻¹ = Wb m⁻²). The direction of F for a positive charge is given by the right-hand rule: point the fingers along v, turn them towards B through the smaller angle, and the thumb gives F. For a negative charge, the force is in the opposite direction.

Facts that follow from the formula:

If the charge is at rest (v = 0), F = 0.
If v is parallel or antiparallel to B (θ = 0° or 180°), F = 0.
If v is perpendicular to B (θ = 90°), F has its maximum value qvB.
F is perpendicular to v, so the force does no work. The kinetic energy and the speed of the charge stay constant. Only the direction of motion changes.

When both an electric field E and a magnetic field B are present, the total force is the Lorentz force: F = q (E + v × B).

Dimension of B: [M T⁻² A⁻¹]. One gauss = 10⁻⁴ tesla.

Motion of a charge in a uniform magnetic field and in combined fields

The table gives the path for each direction of the velocity.

Direction of v	Force	Path
Parallel or antiparallel to B	Zero	Straight line, speed unchanged
Perpendicular to B	qvB, towards the centre	Circle in a plane perpendicular to B
At angle θ (0° < θ < 90°) to B	Acts on the perpendicular component only	Helix (spiral) along B
Charge at rest	Zero	Stays at rest

Circular path. The magnetic force provides the centripetal force: qvB = mv²/r. So

radius r = mv/(qB) = p/(qB), where p is momentum,
angular frequency ω = v/r = qB/m,
period T = 2πm/(qB), frequency ν = qB/(2πm).

T and ν do not depend on the speed or on the radius. A faster particle moves on a larger circle but takes the same time for each turn. Also, r is proportional to the speed for a given particle. A heavier particle or a faster particle has a larger radius. If a particle is accelerated through a potential difference V from rest, then mv²/2 = qV and r = (1/B) √(2mV/q).

Helical path. Resolve v into v∥ (along B) and v⊥ (perpendicular to B). The component v⊥ gives circular motion of radius r = mv⊥/(qB). The component v∥ is unchanged and moves the charge along B. The pitch (distance advanced in one turn) is p = v∥ T = 2πm v∥/(qB).

Velocity selector. Take an electric field E and a magnetic field B perpendicular to each other, and both perpendicular to the beam. A charge moves undeflected when the electric and magnetic forces are equal and opposite: qE = qvB, so

v = E/B.

This speed does not depend on the charge or the mass, and it is the same for positive and negative charges. Charges with other speeds are deflected and are removed by a narrow slit.

#### Magnetic force on a current-carrying conductor

A straight conductor of length l carrying current I in a uniform field B experiences the force

F = I (l × B), with magnitude F = B I l sinθ

where the vector l points along the current. The force is maximum (BIl) when the wire is perpendicular to B and zero when it is parallel to B. The direction is given by the same right-hand rule as for a moving charge, with the current direction taken as the direction of motion of positive charge. This force is the sum of the forces on the moving charge carriers in the wire.

Biot–Savart law and the field of a circular loop

The magnetic field dB at a point at position vector r from a small current element I dl is

dB = (μ₀/4π) I dl × r / r³, so magnitude dB = (μ₀/4π) I dl sinθ / r²

where μ₀ = 4π × 10⁻⁷ T m A⁻¹ is the permeability of free space. The field is proportional to I and to dl and inversely proportional to the square of the distance. Its direction is perpendicular to both dl and r. The field due to a current element is zero along the line of the element (θ = 0).

Circular loop of radius R, N turns, current I.

On the axis at distance x from the centre: B = μ₀ N I R² / [2 (R² + x²)^(3/2)].
At the centre (x = 0): B = μ₀ N I / (2R).
Far away (x ≫ R): B = μ₀ N I R² / (2x³), which falls as 1/x³.

The direction of the field on the axis is given by the right-hand thumb rule: curl the fingers of the right hand along the current in the loop and the thumb points along B. The field lines near the loop are circles around the wire, and near the axis they are almost straight.

Ampere's circuital law and its applications

Ampere's circuital law: the line integral of B around any closed path equals μ₀ times the total current enclosed by the path.

∮ B · dl = μ₀ I(enclosed)

It is useful when the situation has symmetry.

Long straight wire. Take a circle of radius r around the wire. B is constant on the circle and tangent to it, so B (2πr) = μ₀ I and

B = μ₀ I / (2π r).

The field lines are circles around the wire. The direction is given by the right-hand thumb rule: thumb along the current, and the curled fingers show the direction of B.

Long solenoid. For a solenoid with n turns per unit length carrying current I, the field inside (away from the ends) is uniform and parallel to the axis:

B = μ₀ n I

The field outside an ideal long solenoid is very small (taken as zero). The value inside does not depend on the diameter or on the position of the point inside. Use a rectangular Amperian loop with one side inside to get this result.

Force between two parallel currents and the definition of the ampere

Two long parallel wires a distance d apart carry currents I₁ and I₂. The field of wire 1 at wire 2 is μ₀I₁/(2πd). So the force per unit length on each wire is

f = μ₀ I₁ I₂ / (2π d)

Currents in the same direction: the wires attract each other.
Currents in opposite directions: the wires repel each other.
The two forces are equal in size and opposite in direction.

Definition of the ampere. One ampere is the current that, when flowing in each of two infinitely long, thin, parallel wires placed 1 m apart in vacuum, produces a force of 2 × 10⁻⁷ N on each metre of length of each wire.

Torque on a current loop and the magnetic dipole moment

A rectangular loop of sides a and b, with N turns, carries current I in a uniform field B. The net force on the loop is zero. The forces on the two sides form a couple, so there is a torque

τ = N I A B sinθ, where A = ab is the area.

Here θ is the angle between B and the normal to the plane of the loop. The torque is zero when the plane is perpendicular to B (θ = 0°, normal along B) and maximum, NIAB, when the plane is parallel to B (θ = 90°).

The magnetic dipole moment of the loop is m = N I A. It is a vector along the normal to the loop, in the direction given by the right-hand thumb rule. Its unit is A m². Then τ = m × B, with magnitude mB sinθ. A circulating charge or a current loop behaves like a small bar magnet.

Moving coil galvanometer, ammeter and voltmeter

A moving coil galvanometer has a coil of N turns and area A, pivoted between the poles of a permanent magnet with a cylindrical soft iron core. The field is radial, so the plane of the coil is always parallel to B and the torque is NIAB for every deflection. The spring gives a restoring torque kφ, where φ is the deflection and k is the torsional constant. At equilibrium

NIAB = kφ, so φ = (NAB/k) I.

Current sensitivity = φ/I = NAB/k.
Voltage sensitivity = φ/V = NAB/(kR), where R is the resistance of the coil.
Increasing the number of turns raises the current sensitivity, but it also raises the resistance, so the voltage sensitivity may not rise in the same ratio.

Conversion of a galvanometer. Let G be the resistance of the galvanometer and I_g the current for full-scale deflection.

Instrument	Connection	Formula	Resistance of the instrument
Ammeter (range 0 to I)	Shunt S in parallel with G	S = I_g G / (I − I_g)	Less than S and very small
Voltmeter (range 0 to V)	Resistance R in series with G	R = V/I_g − G	Large, G + R

An ammeter is connected in series in the circuit and an ideal ammeter has zero resistance. A voltmeter is connected in parallel across the component and an ideal voltmeter has infinite resistance.

#### Worked examples

Example 1. A proton (m = 1.6 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C) moves at 2 × 10⁵ m s⁻¹ perpendicular to a field of 0.2 T. Find the radius and the period.

r = mv/(qB) = (1.6 × 10⁻²⁷ × 2 × 10⁵)/(1.6 × 10⁻¹⁹ × 0.2) = 1 × 10⁻² m = 1 cm.

T = 2πm/(qB) = 2π × 1.6 × 10⁻²⁷/(1.6 × 10⁻¹⁹ × 0.2) = π × 10⁻⁷ s, about 3.1 × 10⁻⁷ s.

Example 2. A solenoid has 500 turns per metre and carries 4 A. Find the field inside it.

B = μ₀ n I = 4π × 10⁻⁷ × 500 × 4 = 8π × 10⁻⁴ T, about 2.5 × 10⁻³ T.

Example 3. A galvanometer of resistance 90 Ω gives full-scale deflection for 10 mA. Convert it into an ammeter of range 0 to 100 mA, and into a voltmeter of range 0 to 10 V.

Shunt: S = I_g G/(I − I_g) = (10 × 90)/(100 − 10) = 10 Ω, in parallel.

Series resistance: R = V/I_g − G = 10/0.01 − 90 = 1000 − 90 = 910 Ω.

Example 4. A coil of 100 turns and area 0.01 m² carries 2 A and has its plane parallel to a field of 0.5 T. Find the torque.

τ = N I A B = 100 × 2 × 0.01 × 0.5 = 1 N m.

#### Formula summary

Quantity	Formula	Symbols and unit
Force on moving charge	F = qvB sinθ	q in C, v in m s⁻¹, B in T
Radius of circular path	r = mv/(qB)	m in kg, r in m
Period of revolution	T = 2πm/(qB)	T in s
Pitch of helix	p = 2πm v cosθ/(qB)	θ is the angle of v with B
Velocity selector	v = E/B	E in V m⁻¹
Force on a wire	F = BIl sinθ	l in m, I in A
Field of a long wire	B = μ₀I/(2πr)	r in m
Field at centre of a loop	B = μ₀NI/(2R)	R in m
Field in a solenoid	B = μ₀nI	n in turns m⁻¹
Force per length, parallel wires	f = μ₀I₁I₂/(2πd)	N m⁻¹
Torque on a loop	τ = NIAB sinθ = mB sinθ	N m
Magnetic moment	m = NIA	A m²
Shunt	S = I_g G/(I − I_g)	Ω
Series resistance	R = V/I_g − G	Ω

Constants: μ₀ = 4π × 10⁻⁷ T m A⁻¹, e = 1.6 × 10⁻¹⁹ C, mₑ = 9.1 × 10⁻³¹ kg.

NEET traps

1. Magnetic force does no work, so it cannot change the kinetic energy. An electric force can.
2. The force on a charge moving parallel to B is zero, not maximum. The maximum is at 90°.
3. The period T = 2πm/qB does not depend on speed. The radius r = mv/qB does.
4. A negative charge circles in the direction opposite to a positive charge in the same field. The size of the radius is the same for equal |q|, m and v.
5. The velocity selector speed E/B is independent of charge and mass. Particles of other speeds are deflected.
6. Parallel currents in the same direction attract. Parallel charges moving side by side in the same direction do not behave like parallel electric charges, which repel.
7. A shunt is in parallel and small. A voltmeter resistance is in series and large. Do not swap them.
8. An ammeter has less resistance than the galvanometer it was made from. A voltmeter has more resistance.
9. Torque on a loop is maximum when its plane is parallel to B (normal perpendicular to B), and zero when the plane is perpendicular to B.
10. The field inside an ideal solenoid does not depend on the diameter or on the point inside. The field outside is nearly zero.

One-liners

1. Oersted showed that a current produces a magnetic field.
2. The SI unit of magnetic field is the tesla; 1 T = 10⁴ gauss.
3. The magnetic force on a charge at rest is zero.
4. The magnetic force is always perpendicular to velocity.
5. In a uniform field a charge moving perpendicular to B follows a circle.
6. A charge with a component of velocity along B follows a helix.
7. The radius of the circular path is mv/qB.
8. The frequency of revolution is qB/2πm.
9. A velocity selector passes charges of speed E/B.
10. The force on a wire in a field is BIl sinθ.
11. μ₀ = 4π × 10⁻⁷ T m A⁻¹.
12. The field of a long wire falls as 1/r.
13. The field inside a long solenoid is μ₀nI.
14. Currents in opposite directions in parallel wires repel.
15. A current loop of moment m in field B has torque mB sinθ.

Practice questions

A particle of charge q moves with velocity v at an angle θ to a uniform magnetic field B. The magnitude of the magnetic force on it is
1. qvB
2. qv sinθ / B
3. qvB cosθ
4. qvB sinθ
Answer
D. qvB sinθ
F = q v B sinθ, from F = q v × B.
A charged particle of mass m and charge q moves with speed v perpendicular to a uniform magnetic field B. The radius of its circular path is
1. qB/(mv)
2. qvB/m
3. mv/(qB)
4. mv²/(qB)
Answer
C. mv/(qB)
The magnetic force supplies the centripetal force: qvB = mv²/r, so r = mv/(qB).
A proton (mass 1.6 × 10⁻²⁷ kg, charge 1.6 × 10⁻¹⁹ C) moves at 2 × 10⁵ m s⁻¹ perpendicular to a uniform magnetic field of 0.2 T. The radius of its path is
1. 1 cm
2. 2 cm
3. 10 cm
4. 0.5 cm
Answer
A. 1 cm
r = mv/(qB) = (1.6 × 10⁻²⁷ × 2 × 10⁵)/(1.6 × 10⁻¹⁹ × 0.2) = 10⁻² m = 1 cm.
A long straight wire carries a current of 10 A. Using μ₀ = 4π × 10⁻⁷ T m A⁻¹, the magnetic field at a distance of 0.2 m from the wire is
1. 5 × 10⁻⁶ T
2. 1 × 10⁻⁵ T
3. 2 × 10⁻⁵ T
4. 1 × 10⁻⁴ T
Answer
B. 1 × 10⁻⁵ T
B = μ₀I/(2πr) = 2 × 10⁻⁷ × 10/0.2 = 1 × 10⁻⁵ T.
A long solenoid has 2000 turns per metre and carries a current of 0.5 A. Using μ₀ = 4π × 10⁻⁷ T m A⁻¹, the magnetic field inside it, far from the ends, is
1. 8π × 10⁻⁴ T
2. 2π × 10⁻⁴ T
3. 4π × 10⁻⁴ T
4. 4π × 10⁻⁷ T
Answer
C. 4π × 10⁻⁴ T
B = μ₀nI = 4π × 10⁻⁷ × 2000 × 0.5 = 4π × 10⁻⁴ T.
Two long parallel wires 0.1 m apart carry currents of 5 A and 10 A in the same direction. Using μ₀ = 4π × 10⁻⁷ T m A⁻¹, the force per unit length between them is
1. 5 × 10⁻⁵ N m⁻¹
2. 1 × 10⁻³ N m⁻¹
3. 2 × 10⁻⁴ N m⁻¹
4. 1 × 10⁻⁴ N m⁻¹
Answer
D. 1 × 10⁻⁴ N m⁻¹
f = μ₀I₁I₂/(2πd) = 2 × 10⁻⁷ × 5 × 10/0.1 = 1 × 10⁻⁴ N m⁻¹ (attractive).
A circular coil of 20 turns and radius 0.2 m carries a current of 5 A. Using μ₀ = 4π × 10⁻⁷ T m A⁻¹, the magnetic field at its centre is
1. π × 10⁻⁴ T
2. 2π × 10⁻⁴ T
3. (π/2) × 10⁻⁴ T
4. π × 10⁻³ T
Answer
A. π × 10⁻⁴ T
B = μ₀NI/(2R) = 4π × 10⁻⁷ × 20 × 5/(2 × 0.2) = π × 10⁻⁴ T.
A galvanometer of resistance 99 Ω gives full-scale deflection for a current of 10 mA. The shunt resistance needed to convert it into an ammeter of range 0 to 1 A is
1. 99 Ω
2. 1 Ω
3. 0.99 Ω
4. 10 Ω
Answer
B. 1 Ω
S = I_gG/(I − I_g) = (0.01 × 99)/(1 − 0.01) = 0.99/0.99 = 1 Ω. Check: voltage across G is 0.99 V, so the shunt carries 0.99 A at 1 Ω.
A galvanometer of resistance 100 Ω gives full-scale deflection for 5 mA. The series resistance needed to make it a voltmeter of range 0 to 5 V is
1. 1000 Ω
2. 500 Ω
3. 1100 Ω
4. 900 Ω
Answer
D. 900 Ω
R = V/I_g − G = 5/0.005 − 100 = 1000 − 100 = 900 Ω. Check: total resistance 1000 Ω gives 5 V at 5 mA.
A rectangular coil of 50 turns and sides 10 cm × 20 cm carries a current of 2 A. Its plane is parallel to a uniform magnetic field of 0.5 T. The torque on the coil is
1. 0.5 N m
2. 0
3. 1.0 N m
4. 2.0 N m
Answer
C. 1.0 N m
A = 0.02 m². τ = NIAB sinθ with θ = 90° between B and the normal = 50 × 2 × 0.02 × 0.5 = 1.0 N m.
In a region with a uniform electric field of 2 × 10⁴ V m⁻¹ perpendicular to a uniform magnetic field of 0.4 T, a charged particle passes undeflected. Its speed is
1. 5 × 10⁴ m s⁻¹
2. 5 × 10³ m s⁻¹
3. 5 × 10⁵ m s⁻¹
4. 8 × 10³ m s⁻¹
Answer
A. 5 × 10⁴ m s⁻¹
For no deflection qE = qvB, so v = E/B = 2 × 10⁴/0.4 = 5 × 10⁴ m s⁻¹.
A charged particle moves perpendicular to a uniform magnetic field. As it moves, its kinetic energy
1. increases continuously
2. remains constant
3. decreases continuously
4. oscillates between zero and a maximum
Answer
B. remains constant
The magnetic force is perpendicular to the velocity, so it does no work and the speed stays constant.
The time period of a charged particle moving in a circle in a uniform magnetic field depends on
1. its mass, its charge and the magnetic field only
2. its speed only
3. its speed and the radius of the circle
4. its kinetic energy and its charge
Answer
A. its mass, its charge and the magnetic field only
T = 2πm/(qB) contains no speed or radius.
A square loop of side 0.2 m has 10 turns and carries a current of 3 A. The magnitude of its magnetic dipole moment is
1. 6.0 A m²
2. 0.12 A m²
3. 1.2 A m²
4. 0.6 A m²
Answer
C. 1.2 A m²
m = NIA = 10 × 3 × (0.2)² = 1.2 A m².
A wire of length 0.5 m carries a current of 4 A in a uniform magnetic field of 0.3 T. The wire makes an angle of 30° with the field. The force on the wire is
1. 0.52 N
2. 0.6 N
3. 0.15 N
4. 0.3 N
Answer
D. 0.3 N
F = BIl sinθ = 0.3 × 4 × 0.5 × 0.5 = 0.3 N.
A proton and an alpha particle (mass four times, charge twice that of a proton) are accelerated from rest through the same potential difference and enter the same uniform magnetic field perpendicular to it. The ratio of the radius of the proton path to that of the alpha particle path is
1. 1 : 2
2. 1 : √2
3. √2 : 1
4. 2 : 1
Answer
B. 1 : √2
r = (1/B)√(2mV/q), so r ∝ √(m/q). For the proton m/q = 1 unit, for the alpha particle 4/2 = 2 units. Ratio = 1 : √2.
Given below are two statements: Statement I: The magnetic force on a charge at rest in a magnetic field is zero. Statement II: The magnetic force on a charge moving parallel to the magnetic field is zero.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
F = qvB sinθ is zero for v = 0 and for θ = 0°.
Given below are two statements: Statement I: A magnetic force can increase the speed of a charged particle. Statement II: The magnetic force on a moving charge is always perpendicular to its velocity.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
The force is perpendicular to v, so it does no work and cannot change the speed.
Given below are two statements: Statement I: The magnetic field inside a long solenoid is uniform and increases with the diameter of the solenoid. Statement II: The magnetic field outside an ideal long solenoid is nearly zero.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
B = μ₀nI does not depend on the diameter, so Statement I is incorrect. The field outside is nearly zero.
Given below are two statements: Statement I: Two long parallel wires carrying currents in the same direction attract each other. Statement II: Two long parallel wires carrying currents in opposite directions attract each other.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Same direction: attraction. Opposite directions: repulsion.
Given below are two statements: Statement I: The current sensitivity of a moving coil galvanometer is NAB/k. Statement II: The voltage sensitivity is NAB/k and does not depend on the resistance of the coil.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Voltage sensitivity is φ/V = NAB/(kR), which depends on the coil resistance R.
Which of the following statements are correct? 1. An ammeter is made by connecting a small resistance in parallel with a galvanometer. 2. A voltmeter is made by connecting a large resistance in parallel with a galvanometer. 3. An ideal ammeter has zero resistance.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
C. 1 and 3 only
A voltmeter needs a large resistance in series, so statement 2 is wrong. Statements 1 and 3 are correct.
Which of the following statements are correct? 1. The torque on a current loop in a uniform field is zero when its plane is perpendicular to the field. 2. The net force on a current loop in a uniform field is zero. 3. The torque on a current loop is maximum when its plane is perpendicular to the field.
1. 2 and 3 only
2. 1 and 2 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
B. 1 and 2 only
τ = NIAB sinθ with θ measured from B to the normal. It is zero when the plane is perpendicular to B and maximum when the plane is parallel to B. So 3 is wrong.
Given below are two statements: Statement I: One ampere is the current that, in two long parallel wires 1 cm apart in vacuum, gives a force of 2 × 10⁻⁷ N on each metre of wire. Statement II: The force per unit length between two long parallel wires is μ₀I₁I₂/(2πd).
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
The definition uses wires 1 m apart, so Statement I is wrong. Statement II is the correct formula.
Given below are two statements: Statement I: In a velocity selector, charged particles of speed E/B pass undeflected whatever the sign of their charge. Statement II: The undeflected speed in a velocity selector depends on the charge and the mass of the particle.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
v = E/B is independent of q and m, so Statement II is incorrect.
Given below are two statements: Statement I: A charge moving parallel to a uniform magnetic field follows a circular path. Statement II: A charge moving perpendicular to a uniform magnetic field follows a straight line.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
The first moves in a straight line and the second in a circle, so both statements are incorrect.
Assertion (A): A charged particle moving perpendicular to a uniform magnetic field has constant kinetic energy. Reason (R): The magnetic force on the particle is perpendicular to its velocity and does no work.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
No work is done, so kinetic energy is constant. R explains A.
Assertion (A): The frequency of revolution of a charged particle in a uniform magnetic field does not depend on its speed. Reason (R): The magnetic force on a charge is perpendicular to its velocity.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Frequency is qB/(2πm), independent of speed. R is true but the reason for speed independence is that r ∝ v, not the direction of the force alone.
Assertion (A): A current loop in a uniform magnetic field experiences a torque but no net force. Reason (R): The forces on opposite sides of the loop are equal and opposite but do not act along the same line.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Equal and opposite forces give zero net force. Their separate lines of action give a couple.
Assertion (A): A shunt resistance connected in parallel with a galvanometer converts it into an ammeter. Reason (R): The shunt has a very high resistance compared with the galvanometer.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
The shunt has a very small resistance, so R is false.
Assertion (A): A stationary charge experiences a magnetic force in a magnetic field. Reason (R): The magnetic force on a charge depends on its velocity.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
F = qvB sinθ is zero at v = 0, so A is false; R is true.
Match List I with List II. List I: a. Radius of circular path of a charge in a magnetic field b. Time period of revolution c. Field inside a long solenoid d. Field at distance r from a long straight wire. List II: i. μ₀nI ii. μ₀I/(2πr) iii. mv/(qB) iv. 2πm/(qB)
1. a-iii, b-iv, c-i, d-ii
2. a-iii, b-iv, c-ii, d-i
3. a-i, b-iv, c-iii, d-ii
4. a-iv, b-iii, c-i, d-ii
Answer
A. a-iii, b-iv, c-i, d-ii
Radius mv/qB; period 2πm/qB; solenoid μ₀nI; wire μ₀I/2πr.
Match List I with List II. List I: a. Magnetic field b. Magnetic dipole moment c. Permeability of free space d. Torque. List II: i. A m² ii. T m A⁻¹ iii. Wb m⁻² iv. N m
1. a-i, b-iii, c-ii, d-iv
2. a-iii, b-i, c-ii, d-iv
3. a-iii, b-i, c-iv, d-ii
4. a-ii, b-i, c-iii, d-iv
Answer
B. a-iii, b-i, c-ii, d-iv
B is in Wb m⁻² (tesla); m in A m²; μ₀ in T m A⁻¹; torque in N m.
Match List I with List II. List I: a. Ammeter b. Voltmeter c. Velocity selector d. Moving coil galvanometer. List II: i. Crossed electric and magnetic fields ii. High resistance in series iii. Low resistance in parallel iv. Radial magnetic field and a pivoted coil
1. a-iii, b-ii, c-iv, d-i
2. a-ii, b-iii, c-i, d-iv
3. a-i, b-ii, c-iii, d-iv
4. a-iii, b-ii, c-i, d-iv
Answer
D. a-iii, b-ii, c-i, d-iv
Ammeter: shunt in parallel. Voltmeter: series resistance. Velocity selector: crossed E and B. Galvanometer: radial field.
Match List I with List II. List I: a. Velocity parallel to B b. Velocity perpendicular to B c. Velocity at an angle between 0° and 90° to B d. Charge at rest. List II: i. Helix ii. Stays at rest iii. Circle iv. Straight line
1. a-iv, b-iii, c-i, d-ii
2. a-iv, b-i, c-iii, d-ii
3. a-iii, b-iv, c-i, d-ii
4. a-ii, b-iii, c-i, d-iv
Answer
A. a-iv, b-iii, c-i, d-ii
Parallel: no force, straight line. Perpendicular: circle. Oblique: helix. At rest: no force.
Match List I with List II. A straight wire of length L carries current I in a uniform field B. List I: a. L = 1 m, I = 2 A, B = 0.5 T, wire perpendicular to B b. Same wire at 30° to B c. Same wire parallel to B d. L = 2 m, I = 2 A, B = 0.5 T, wire perpendicular to B. List II: i. 0.5 N ii. 2 N iii. 0 iv. 1 N
1. a-iv, b-iii, c-i, d-ii
2. a-i, b-iv, c-iii, d-ii
3. a-iv, b-i, c-iii, d-ii
4. a-iv, b-i, c-ii, d-iii
Answer
C. a-iv, b-i, c-iii, d-ii
F = BIL sinθ: (a) 0.5 × 2 × 1 = 1 N; (b) 0.5 N; (c) 0; (d) 0.5 × 2 × 2 = 2 N.
Match List I with List II. List I: a. Oersted's experiment b. Biot–Savart law c. Ampere's circuital law d. Lorentz force. List II: i. Line integral of B around a closed path equals μ₀ times the enclosed current ii. A compass needle is deflected by a current-carrying wire iii. Force on a charge in both electric and magnetic fields iv. Field of a small current element
1. a-ii, b-iv, c-iii, d-i
2. a-i, b-iv, c-ii, d-iii
3. a-iv, b-ii, c-i, d-iii
4. a-ii, b-iv, c-i, d-iii
Answer
D. a-ii, b-iv, c-i, d-iii
Direct matching of each name to its statement.
A charged particle moving in a uniform magnetic field has a velocity component along the field of 3 × 10⁵ m s⁻¹ and goes round a circle in 2 × 10⁻⁷ s. The pitch of its helical path is
1. 0.06 m
2. 0.03 m
3. 0.6 m
4. 0.12 m
Answer
A. 0.06 m
Pitch = v∥T = 3 × 10⁵ × 2 × 10⁻⁷ = 0.06 m.
An ion of mass 2 × 10⁻²⁶ kg and charge 1.6 × 10⁻¹⁹ C moves on a circle of radius 0.5 m in a uniform magnetic field of 0.4 T. Its speed is
1. 8 × 10⁵ m s⁻¹
2. 1.6 × 10⁶ m s⁻¹
3. 3.2 × 10⁶ m s⁻¹
4. 1.6 × 10⁷ m s⁻¹
Answer
B. 1.6 × 10⁶ m s⁻¹
v = qBr/m = 1.6 × 10⁻¹⁹ × 0.4 × 0.5/(2 × 10⁻²⁶) = 1.6 × 10⁶ m s⁻¹.
A galvanometer of resistance 100 Ω shows full-scale deflection for 2 mA. To read up to 20 V, it should be connected in series with a resistance of
1. 900 Ω
2. 10100 Ω
3. 9900 Ω
4. 10000 Ω
Answer
C. 9900 Ω
R = V/I_g − G = 20/0.002 − 100 = 10000 − 100 = 9900 Ω.
A galvanometer of resistance 45 Ω is shunted so that only one tenth of the total current passes through it. The shunt resistance is
1. 0.5 Ω
2. 5 Ω
3. 4.5 Ω
4. 9 Ω
Answer
B. 5 Ω
Galvanometer current = I/10, shunt current = 9I/10. Equal voltage: (I/10) × 45 = (9I/10) × S, so S = 5 Ω.
Two long parallel wires 10 cm apart carry currents of 6 A and 4 A in opposite directions. Using μ₀ = 4π × 10⁻⁷ T m A⁻¹, the force per unit length between them is
1. 4.8 × 10⁻⁵ N m⁻¹, repulsive
2. 2.4 × 10⁻⁵ N m⁻¹, repulsive
3. 4.8 × 10⁻⁵ N m⁻¹, attractive
4. 9.6 × 10⁻⁵ N m⁻¹, attractive
Answer
A. 4.8 × 10⁻⁵ N m⁻¹, repulsive
f = 2 × 10⁻⁷ × 6 × 4/0.1 = 4.8 × 10⁻⁵ N m⁻¹. Opposite currents repel.
Two long straight parallel wires 0.2 m apart each carry 5 A, in opposite directions. Using μ₀ = 4π × 10⁻⁷ T m A⁻¹, the magnetic field at the midpoint between them is
1. 1 × 10⁻⁵ T
2. 0
3. 4 × 10⁻⁵ T
4. 2 × 10⁻⁵ T
Answer
D. 2 × 10⁻⁵ T
Each wire gives 2 × 10⁻⁷ × 5/0.1 = 1 × 10⁻⁵ T at the midpoint. For opposite currents the two fields point the same way between the wires, so they add: 2 × 10⁻⁵ T.
A coil of 200 turns and area 5 × 10⁻⁴ m² carries 0.1 A in a radial magnetic field of 0.2 T, so that its plane is always parallel to the field. The torque on it is
1. 1 × 10⁻³ N m
2. 2 × 10⁻³ N m
3. 4 × 10⁻³ N m
4. 2 × 10⁻⁵ N m
Answer
B. 2 × 10⁻³ N m
τ = NIAB = 200 × 0.1 × 5 × 10⁻⁴ × 0.2 = 2 × 10⁻³ N m.

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1.The work done in moving a charge +Q across two points having a potential difference K is:
2.What is the work done in moving a unit positive charge from infinity to that point, at that point?
3.What can be the shape of the trajectory of a charged particle moving in a uniform magnetic field?
4.If 90 J of work is done in moving a charge of 2,000 coulombs across V volts, find V.
5.If 24 joules of work is done in moving a charge of 2 coulombs across two points, then what will be the correct voltage across these two points?
6.If 36 J of work is done in moving a charge Q across 8 V, find Q (in coulombs).
7.Find the work done (in joules) in moving a charge of 1.6 coulombs across a potential difference of 0.8 V.
8.How much work is done in moving a charge of 4 C across two points having a potential difference of 12 V?
9.How much work is done in moving a charge of 10 C across two points having a potential difference 20 V?
10.How much work (W) is done in moving a charge (Q) of 5 C across two points that have a potential difference of 15 V?

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