An electron enters a region of uniform magnetic field with velocity v perpendicular to the field. If the magnetic field strength is B, the radius of curvature is:
Answer: A
For a charged particle in a magnetic field, centripetal force equals magnetic force: mv²/r = evB, giving r = mv/eB. This is the radius of the circular path.
Q.2Medium
Two parallel wires carrying currents I₁ and I₂ in opposite directions are separated by distance d. The force between them is:
Answer: C
Currents in opposite directions repel each other. The force per unit length is F/ℓ = μ₀I₁I₂/2πd, making total force F = μ₀I₁I₂ℓ/2πd (repulsive).
Q.3Medium
A rectangular loop of dimensions a × b carrying current I is placed in a uniform magnetic field B. The maximum torque on the loop is:
Answer: A
Torque on a current loop in a magnetic field is τ = NIAB sin(θ), where A is the area and θ is the angle. Maximum torque occurs when sin(θ) = 1, giving τ_max = BIab for N=1.
Q.4Medium
The magnetic moment of an electron orbiting in the first Bohr orbit is approximately:
Answer: A
The magnetic moment of an electron in the first Bohr orbit equals 1 Bohr magneton (μ_B = eℏ/2m_e ≈ 9.27 × 10⁻²⁴ J/T). This is a fundamental quantum result.
Q.5Medium
The magnetic field due to a long straight wire carrying current I at perpendicular distance r is:
Answer: A
Using Ampere's law for a long straight wire, ∮B·dl = μ₀I_enclosed. For a circular path of radius r: B(2πr) = μ₀I, giving B = μ₀I/2πr.
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Q.6Medium
A proton and an electron, both accelerated through the same potential difference, enter a uniform magnetic field perpendicularly. Which has a larger radius of curvature?
Answer: A
Both particles gain same kinetic energy, so mv²/2 is same. Since r = mv/eB and proton has much larger mass than electron, proton has larger radius of curvature.
Q.7Medium
The period of revolution of a charged particle in a magnetic field is independent of:
Answer: C
Period T = 2πm/eB is independent of velocity. This remarkable result means all particles with same m and q have same period regardless of speed in a given B field.
Q.8Medium
A compass needle placed in a magnetic field experiences a maximum torque when the needle is:
Answer: C
Torque τ = m × B has magnitude τ = mB sin(θ). Maximum occurs when sin(θ) = 1, i.e., θ = 90° (perpendicular orientation).
Q.9Medium
The self-inductance of a solenoid with N turns, length L, and cross-sectional area A is:
Answer: A
Self-inductance of solenoid is derived from L = NΦ/I where Φ = μ₀nIA. This gives L = μ₀N²A/L, proportional to N² and inversely proportional to length.
Q.10Medium
A solenoid with 500 turns is 0.5 m long and carries a current of 2 A. The permeability of free space is μ₀ = 4π × 10⁻⁷ T·m/A. Calculate the magnetic field inside the solenoid.
Answer: C
B = μ₀nI where n = N/L = 0500.5 = 1000 turns/m. B = 4π × 10⁻⁷ × 1000 × 2 = 2.51 × 10⁻¹ T ≈ 0.251 T.
Q.11Medium
Two parallel wires carry currents I₁ = 5 A and I₂ = 3 A in the same direction, separated by distance r = 0.1 m. The force per unit length between them is approximately:
A rectangular loop ABCD with sides 2 m × 3 m carries a current of 4 A and is placed in a uniform magnetic field of 0.5 T perpendicular to the plane of the loop. The magnetic torque on the loop is:
Answer: A
Torque τ = NIAB sin θ. When B is perpendicular to the plane of the loop, it is parallel to the normal of the loop area, so θ = 0° and τ = 0.
Q.13Medium
A long straight wire carries a current and produces a magnetic field. At a distance of 2 cm from the wire, the field is 4 × 10⁻⁵ T. What is the current in the wire? (μ₀ = 4π × 10⁻⁷ T·m/A)
Answer: A
B = μ₀I/(2πr). So I = 2πrB/μ₀ = 2π × 0.02 × 4 × 10⁻⁵/(4π × 10⁻⁷) = 2 A.
Q.14Medium
A beam of electrons is accelerated through a potential difference V and then enters a region of perpendicular electric and magnetic fields. For the electrons to move undeflected, which condition must be satisfied?
Answer: B
For undeflected motion, electric force equals magnetic force: qE = qvB, which simplifies to E = vB. This is the velocity selector condition.
Q.15Medium
A conducting rod of length L = 0.5 m moves with velocity v = 10 m/s perpendicular to a uniform magnetic field B = 2 T. The motional EMF induced is:
Answer: B
Motional EMF ε = BLv = 2 × 0.5 × 10 = 10 V.
Q.16Medium
In a cyclotron, a charged particle spirals outward as it gains energy. The frequency of revolution is independent of:
Answer: B
The cyclotron frequency f = qB/(2πm) is independent of velocity. As velocity increases, radius increases but period remains constant (independent of v).
Q.17Medium
A rectangular conducting loop is partially inside a uniform magnetic field region. If the loop is pulled out with constant velocity v, the induced EMF depends on:
Answer: A
EMF induced ε = BLv where L is the length of the conductor cutting magnetic field lines and v is the velocity. It depends on all three factors: B, L, and v.
Q.18Medium
A rectangular coil of area A and N turns is rotated in a uniform magnetic field B with angular velocity ω. The maximum induced EMF is:
Answer: A
Maximum EMF in rotating coil: ε_max = NABω, where Φ = NBA cos(ωt), so ε = dΦ/dt = NABω sin(ωt)
Q.19Medium
A solenoid of length L and cross-sectional area A has N turns. Its self-inductance is:
Answer: A
Self-inductance of solenoid: L = μ₀n²V = μ₀(N/L)²AL = μ₀N²A/L
Q.20Medium
A conducting rod of mass m slides on two parallel rails separated by distance d in a magnetic field B perpendicular to the plane. If the rod moves with constant velocity v, the magnetic force on it is:
Answer: B
Induced EMF: ε = Bdv, induced current: I = Bdv/R, magnetic force: F = BId = B²d²v/R (opposes motion)
