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.
A straight wire carrying current I is placed in a uniform magnetic field B at an angle θ to the magnetic field. If the length of the wire is L, the magnetic force on the wire is:
Answer: B
The magnetic force on a current-carrying conductor is F = BIL sinθ, where θ is the angle between the current direction and the magnetic field. When θ = 90°, force is maximum (BIL), and when θ = 0°, force is zero.
Q.462Easy
Two parallel wires carry currents I₁ and I₂ in the same direction, separated by distance d. The force per unit length between them is:
Answer: B
Parallel wires carrying currents in the same direction experience attractive force. Force per unit length = μ₀I₁I₂/(2πd). If currents are opposite, the force is repulsive.
Q.463Easy
A charged particle enters a uniform magnetic field with velocity perpendicular to the field. The particle will move in:
Answer: B
When a charged particle moves perpendicular to a uniform magnetic field, the Lorentz force acts as centripetal force, causing circular motion. The radius is r = mv/(qB).
Q.464Easy
A solenoid has n turns per unit length and carries current I. The magnetic field inside the solenoid is:
Answer: A
The magnetic field inside an ideal long solenoid is B = μ₀nI, independent of the solenoid's radius and position along the axis (away from ends). This assumes n is the number of turns per unit length.
Q.465Easy
The magnetic moment of a current loop is defined as:
Answer: A
Magnetic moment M = IA, where I is the current and A is the area enclosed by the loop. For N turns, M = NIA. The SI unit is A·m².
Q.466Easy
An electron moves in a plane perpendicular to a uniform magnetic field. If the radius of its circular path is r, the momentum of the electron is:
Answer: A
For circular motion in a magnetic field: qvB = mv²/r, which gives r = mv/(qB). Therefore, momentum p = mv = qBr = eBr (for an electron where q = e).
Q.467Medium
The magnetic field between the poles of a permanent magnet is approximately:
Answer: B
Between the parallel poles of a strong permanent magnet (in the central region), the magnetic field is approximately uniform. However, it becomes non-uniform near the pole edges.
Q.468Medium
A rectangular conducting loop ABCD with sides a and b is rotated with angular velocity ω in a uniform magnetic field B perpendicular to the plane of rotation. The induced EMF is:
Answer: B
When the loop rotates, the magnetic flux through it varies as Φ = BA·cosωt. The induced EMF = -dΦ/dt = BA·ω·sinωt = Bab·ω·sinωt
Q.469Medium
Two magnets are placed with their north poles facing each other. The force between them varies with distance r as:
Answer: D
Two magnetic dipoles interact with force F ∝ 1/r⁴ when aligned along the same axis. This is because the magnetic field of a dipole varies as 1/r³, and force on a dipole is proportional to the field gradient.
Q.470Medium
A conducting rod of length L moves with velocity v perpendicular to its length in a magnetic field B. The motional EMF induced is maximum when:
Answer: B
Motional EMF = B·L·v·sinθ, where θ is the angle between v and B. EMF is maximum when sinθ = 1, i.e., when v is perpendicular to B.
Q.471Medium
The magnetic susceptibility of a paramagnetic material is:
Answer: B
Paramagnetic materials have positive but small magnetic susceptibility (χ > 0, typically 10⁻⁵ to 10⁻³). Diamagnetic materials have small negative susceptibility, and ferromagnetic materials have large positive susceptibility.
Q.472Medium
A charged particle with charge q and mass m is moving with speed v in a circular path of radius r in a magnetic field. The magnetic field strength is:
Answer: A
From qvB = mv²/r (centripetal force equals magnetic force), we get B = mv/(qr). This is the relationship between field strength, particle properties, and circular path radius.
Q.473Medium
The permeability of free space μ₀ has the value:
Answer: B
The permeability of free space μ₀ = 4π × 10⁻⁷ T·m/A. Option A is permittivity ε₀, option C is speed of light, and option D is Planck's constant.
Q.474Medium
A long straight wire carrying current I produces a magnetic field at distance r. If the current is doubled and distance is halved, the magnetic field becomes:
Answer: B
B = μ₀I/(2πr). If I → 2I and r → r/2, then B_new = μ₀(2I)/(2π(r/2)) = 4·μ₀I/(2πr) = 4B_initial
Q.475Medium
The SI unit of magnetic flux density (magnetic field) is:
Answer: B
The SI unit of magnetic field (flux density) is Tesla (T). 1 T = 1 Wb/m² = 1 kg/(A·s²). Weber is the unit of magnetic flux, Gauss is CGS unit, and Henry is unit of inductance.
Q.476Hard
A proton and an alpha particle (He²⁺ nucleus) are accelerated through the same potential difference. They are then made to move perpendicular to a uniform magnetic field. The ratio of their radii of curvature is:
Answer: C
From qVB = mv²/2 and r = mv/(qB), we get r = √(2mV/q)/B. For proton (m=m_p, q=e) and alpha (m=4m_p, q=2e): r_p/r_α = √(m_p/(4m_p))·√(2e/e) = √(41)·√2 = √(21)·√2 = 12
Q.477Hard
A rectangular loop of dimensions a × b is placed in a non-uniform magnetic field where B varies as B = B₀(1 + kx), where x is the distance from a reference line. The net force on the loop is:
Answer: C
In a non-uniform field, the forces on opposite sides of the loop are unequal. The net force depends on the field gradient. F = I·∫(dB/dx)·dA = I·b·∫B₀k·da = B₀kIab (approximately, for small variations).
Q.478Hard
A charged particle enters a region with perpendicular electric and magnetic fields with velocity v. For the particle to pass undeflected, the condition is:
Answer: A
For undeflected motion, electric force equals magnetic force: qE = qvB, which gives E = vB. This is the principle of a velocity selector.
Q.479Hard
A solenoid with N turns, length L, and cross-sectional area A is wound with wire of resistance R. When connected to a voltage source V, the magnetic energy stored is:
Answer: C
Current I = V/R. Self-inductance L = μ₀N²A/L. Magnetic energy = LI²/2 = (μ₀N²A/L)·(V²/R²)/2 = V²μ₀N²A/(2R²L)
Q.480Easy
A light ray travels from air into glass with refractive index 1.5. If the angle of incidence is 30°, what is the angle of refraction?
