Physics is where JEE ranks are usually won or lost, since a single conceptual slip changes the whole answer. Questions here span mechanics, rotational motion, thermodynamics, waves and oscillations, electrostatics, current electricity, magnetism, optics, and modern physics. Numerical problems carry full derivations so you can see which step you skipped, not just which option was right.
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.
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)
