1. A charged particle moves with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$ . The magnetic force experienced by the particle is zero if:

(a) $\vec{v}$ is perpendicular to $\vec{B}$
(b) $\vec{v}$ is parallel to $\vec{B}$
(c) Particle is stationary
(d) Both (b) and (c)

Answer: (d)
$F = qvB \sin\theta$ . Force is zero if $v=0$ (stationary) or $\theta=0^\circ/180^\circ$ (parallel/antiparallel).

2. An electron enters a magnetic field at right angles to it as shown in a diagram. The direction of force acting on the electron will be:

(a) To the right
(b) To the left
(c) Into the page
(d) Out of the page

Answer: (Depends on diagram, usually Into/Out)
Using Fleming’s Left Hand Rule: For an electron (negative charge), reverse the direction of current. If $v$ is right and $B$ is up, Force is Into the page.

3. Two parallel wires carrying currents in the same direction attract each other because of:

(a) Potential difference
(b) Mutual inductance
(c) Magnetic force
(d) Electric force

Answer: (c)
The magnetic field produced by one wire exerts a Lorentz force on the electrons in the other wire.

4. A galvanometer acting as a voltmeter should have:

(a) Low resistance in series
(b) High resistance in series
(c) Low resistance in parallel
(d) High resistance in parallel

Answer: (b)
To convert a galvanometer to a voltmeter, a high resistance is connected in series to limit current and increase voltage range.

5. The magnetic field inside a long straight solenoid carrying current:

(a) Is zero
(b) Decreases as we move towards the ends
(c) Increases as we move towards the ends
(d) Is the same at all points

Answer: (d)
Inside a long ideal solenoid, the field is uniform ( $B = \mu_0 n I$ ). Note: At the exact ends, it drops to $B/2$ .

6. Biot-Savart law indicates that the moving electrons (velocity $v$ ) produce a magnetic field $B$ such that:

(a) $B \perp v$
(b) $B \parallel v$
(c) It obeys inverse cube law
(d) It is along the line joining the electron and observation point

Answer: (a)
$d\vec{B} \propto d\vec{l} \times \vec{r}$ . Since current direction is related to $v$ , $B$ is perpendicular to $v$ .

7. A circular coil of radius $R$ carries a current $I$ . The magnetic field at its center is $B$ . At what distance from the center on the axis is the magnetic field $B/8$ ?

(a) $\sqrt{3}R$
(b) $\sqrt{2}R$
(c) $2R$
(d) $3R$

Answer: (a)
$B_{axis} = \frac{B_{center}}{(1 + x^2/R^2)^{3/2}}$ . If $B_{axis} = B/8$ , then $(1 + x^2/R^2)^{3/2} = 8$ . Solving gives $x = \sqrt{3}R$ .

8. If the number of turns in a moving coil galvanometer is doubled, the current sensitivity:

(a) Remains same
(b) Becomes double
(c) Becomes half
(d) Becomes four times

Answer: (b)
Current sensitivity $I_s = \frac{NBA}{k}$ . If $N$ doubles, $I_s$ doubles.

9. The SI unit of magnetic field is Tesla (T). 1 Tesla is equivalent to:

(a) $1 \text{ N A}^{-1} \text{ m}^{-1}$
(b) $1 \text{ N A m}^{-1}$
(c) $1 \text{ Wb m}^{-1}$
(d) $1 \text{ Gauss}$

Answer: (a)
From $F = IlB$ , $B = F/Il = \text{N} / (\text{A m})$ .

10. An alpha particle and a proton have the same kinetic energy. They enter a uniform magnetic field perpendicular to it. The ratio of their radii ( $R_\alpha : R_p$ ) is:

(a) 1:1
(b) 2:1
(c) 1:2
(d) 4:1

Answer: (a)
$R = \frac{\sqrt{2mK}}{qB}$ . Ratio $\propto \frac{\sqrt{m}}{q}$ .
For proton: $\frac{\sqrt{1}}{1} = 1$ . For alpha: $\frac{\sqrt{4}}{2} = 1$ . Ratio is 1:1.

Part 2: Assertion-Reason Questions

(A) Both A & R are true, R explains A.
(B) Both A & R are true, R does NOT explain A.
(C) A is true, R is false.
(D) A is false, R is true.

1. Assertion (A): Magnetic force on a moving charge is always perpendicular to the magnetic field.
Reason (R): Magnetic force is given by $\vec{F} = q(\vec{v} \times \vec{B})$ .

Answer: (A)
The cross product ensures the resulting force vector is perpendicular to both $\vec{v}$ and $\vec{B}$ .

2. Assertion (A): If a charged particle moves in a circular path in a magnetic field, its kinetic energy remains constant.
Reason (R): The magnetic force acts perpendicular to velocity, so work done is zero.

Answer: (A)
Since $W = \Delta K$ and work is zero, kinetic energy change is zero.

3. Assertion (A): To increase the range of an ammeter, we connect a suitable high resistance in series.
Reason (R): The ammeter must have very low resistance to measure current accurately.

Answer: (D)
Assertion is False. To increase ammeter range, we connect a low resistance (shunt) in parallel. Reason is True.

4. Assertion (A): The magnetic field at the ends of a very long current-carrying solenoid is half of that at the center.
Reason (R): If the solenoid is sufficiently long, the field is uniform everywhere.

Answer: (C)
Assertion is True ( $B_{end} = \frac{1}{2} \mu_0 n I$ ). Reason is False (field drops near ends).

5. Assertion (A): An electron and a proton enter a magnetic field with the same momentum. They describe paths of the same radius.
Reason (R): The radius of the path is given by $r = p/qB$ .

Answer: (A)
Since magnitude of charge $q$ is same for both ( $e$ ) and $p$ is same, $r$ is same.

6. Assertion (A): A current loop placed in a uniform magnetic field behaves like a magnetic dipole.
Reason (R): It experiences a net force but no torque.

Answer: (C)
Assertion is True. Reason is False—in a uniform field, it experiences no net force but does experience torque.

7. Assertion (A): The sensitivity of a moving coil galvanometer is increased by using a soft iron core.
Reason (R): Soft iron has high magnetic permeability and increases the magnetic field strength.

Answer: (A)
Correct. Stronger B means larger deflection for same current ( $\phi \propto B$ ).

8. Assertion (A): Cyclotron is not used to accelerate electrons.
Reason (R): Electrons have very small mass, so they quickly reach relativistic speeds, throwing them out of resonance.

Answer: (A)
Relativistic mass increase changes the frequency of revolution, violating the cyclotron condition.

9. Assertion (A): Parallel currents repel and antiparallel currents attract.
Reason (R): Forces between current carrying wires obey Newton’s Third Law.

Answer: (D)
Assertion is False. Parallel currents attract. Reason is True (forces are equal and opposite).

10. Assertion (A): Magnetic field lines are continuous closed loops.
Reason (R): Magnetic monopoles do not exist.

Answer: (A)
Unlike electric field lines which start/end on charges, magnetic lines have no starting or ending point.

Part 3: Important Derivations & Theory

1. Using Biot-Savart Law, derive the expression for the magnetic field on the axis of a circular current loop. (5 Marks)

Key Steps:
1. Diagram: Loop in y-z plane, point P on x-axis.
2. Element $dl$ : Magnetic field $dB = \frac{\mu_0 I dl}{4\pi r^2}$ perpendicular to $r$ .
3. Resolution: $dB$ has components $dB_x$ (axial) and $dB_\perp$ (vertical). Vertical components cancel.
4. Integration: $B = \int dB_x = \int dB \cos\theta$ .
5. Substitution: $\cos\theta = R/r$ and $r = \sqrt{R^2+x^2}$ .
6. Final Result: $B = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$ .

2. Derive the expression for the force between two parallel long straight conductors carrying currents in the same direction. Hence define 1 Ampere. (3 Marks)

1. Field $B_1$ due to wire 1 at wire 2: $B_1 = \frac{\mu_0 I_1}{2\pi d}$ .
2. Force on length $L$ of wire 2: $F_{21} = I_2 L B_1 \sin 90^\circ$ .
3. Force per unit length: $f = \frac{\mu_0 I_1 I_2}{2\pi d}$ .
Definition: 1 Ampere is that current which, if maintained in two infinite parallel wires 1m apart in vacuum, produces a force of $2 \times 10^{-7}$ N/m.

3. Deduce the expression for the torque acting on a rectangular current loop placed in a uniform magnetic field. (3 Marks)

1. Forces on arms parallel to axis cancel out.
2. Forces on vertical arms are $F = IbB$ , acting in opposite directions (Couple).
3. Perpendicular distance between forces $= b \sin\theta$ (where $\theta$ is angle between area vector and B).
4. Torque $\tau = F \times \text{distance} = IbB \times a \sin\theta = I(ab)B \sin\theta$ .
5. Result: $\tau = mB \sin\theta$ or $\vec{\tau} = \vec{m} \times \vec{B}$ .

Part 4: Numericals

1. A circular coil of wire consisting of 100 turns, each of radius 8.0 cm, carries a current of 0.40 A. What is the magnitude of the magnetic field B at the center of the coil?

Formula: $B = \frac{\mu_0 N I}{2R}$ .
$B = \frac{4\pi \times 10^{-7} \times 100 \times 0.40}{2 \times 0.08}$ .
$B = \pi \times 10^{-4} \approx 3.14 \times 10^{-4}$ T.

2. A galvanometer with a coil resistance of $12 \Omega$ shows full scale deflection for a current of $2.5 \text{ mA}$ . How will you convert it into a voltmeter of range $0$ to $10 \text{ V}$ ?

Given: $G = 12 \Omega$ , $I_g = 2.5 \times 10^{-3} \text{ A}$ , $V = 10 \text{ V}$ .
Series Resistance $R = \frac{V}{I_g} - G$ .
$R = \frac{10}{2.5 \times 10^{-3}} - 12 = 4000 - 12 = 3988 \Omega$ .
Answer: Connect $3988 \Omega$ in series.

3. A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid?

Formula: $B = \mu_0 n I$ , where $n = N/L$ .
$n = 500 / 0.5 = 1000$ turns/m.
$B = 4\pi \times 10^{-7} \times 1000 \times 5 = 20\pi \times 10^{-4} = 6.28 \times 10^{-3}$ T.

Part 5: Case Study

Case Study: Moving Coil Galvanometer (MCG)
The Moving Coil Galvanometer is a device used to detect and measure small electric currents. It is based on the principle that a current-carrying coil placed in a magnetic field experiences a torque. The coil is wound on a light metallic frame and suspended between two cylindrical pole pieces of a strong magnet. A soft iron core is placed inside the coil to make the field radial.

Why is a radial magnetic field used in a galvanometer?
What is the function of the soft iron core?
Write the expression for Current Sensitivity.
How can the current sensitivity be increased?

1. To ensure the plane of the coil is always parallel to field lines, keeping $\theta = 90^\circ$ and torque maximum ( $\tau = NIAB$ ). This makes the scale linear.
2. It increases the strength of the magnetic field and helps in making it radial.
3. $I_s = \frac{\alpha}{I} = \frac{NBA}{k}$ (where k is torsional constant).
4. By increasing Number of turns ( $N$ ), Area ( $A$ ), Magnetic field ( $B$ ), or decreasing torque constant ( $k$ ) using a soft suspension wire like Phosphor-Bronze.

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Moving Charges & Magnetism: Question Bank

Moving Charges & Magnetism: Question Bank

Part 1: Multiple Choice Questions (1 Mark)

Part 2: Assertion-Reason Questions

Part 3: Important Derivations & Theory

Part 4: Numericals

Part 5: Case Study

Editorial Team

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Part 1: Multiple Choice Questions (1 Mark)

Part 2: Assertion-Reason Questions

Part 3: Important Derivations & Theory

Part 4: Numericals

Part 5: Case Study

Editorial Team

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