A circular loop and a square loop of equal perimeter are placed in the same uniform magnetic field. Their magnetic moments are:
Answer: B
For equal perimeter, circle encloses maximum area (isoperimetric inequality). Since magnetic moment m = IA, circular loop with larger area has larger magnetic moment.
Q.2Hard
A charged particle moves in crossed electric and magnetic fields. For the particle to move undeflected, the condition is:
Answer: A
For undeflected motion, electric and magnetic forces must balance: qE = qvB, giving E = vB. This is the principle of velocity selector used in mass spectrometers.
Q.3Hard
The magnetic field inside a toroid with N turns, major radius R, and carrying current I is:
Answer: D
In a toroid, using Ampere's law on circular path of radius r (inside toroid): B(2πr) = μ₀NI, so B = μ₀NI/2πr. Field varies inversely with distance from toroid center.
Q.4Hard
Two identical coils are placed coaxially with separation much larger than their radius. Their mutual inductance is:
Answer: B
For coaxial coils with large separation d >> radius, mutual inductance M ∝ 1/d² due to spreading of magnetic field lines. This is used in wireless power transfer systems.
Q.5Hard
The Hall effect in semiconductors is used to determine:
Answer: C
Hall voltage V_H = BId/ne·t indicates carrier sign from voltage polarity and carrier density n from magnitude. This dual information makes Hall effect powerful for semiconductor characterization.
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Q.6Hard
The magnetic field at the center of a circular arc of radius R subtending angle θ at the center and carrying current I is:
Answer: A
For a circular arc, B = (μ₀I/4πR) × θ, where θ is in radians. This is derived from the Biot-Savart law integrated over the arc.
Q.7Hard
A toroidal magnetic field is produced by a toroid with N turns carrying current I. If the mean radius of the toroid is R and the cross-sectional area of the core is A, the magnetic energy stored is:
Answer: A
Magnetic energy U = ½LI² where L = μ₀N²A/(2πR) for a toroid. Therefore U = μ₀N²I²A/(4πR). Note: The correct formula is actually U = ½ × μ₀N²I²A/(2πR) = μ₀N²I²A/(4πR).
Q.8Hard
An alpha particle (charge +2e, mass 4u) and a proton (charge +e, mass u) are accelerated from rest through the same potential difference. They then enter a uniform magnetic field perpendicular to their motion. The ratio of their radii of curvature is:
Answer: C
After acceleration: ½m₁v₁² = q₁V and ½m₂v₂² = q₂V. In magnetic field, r = mv/(qB). r₁/r₂ = (m₁v₁/q₁)/(m₂v₂/q₂) = (4u × √(2eV/4u)/2e)/(u × √(2eV/u)/e) = √(2u/e) × e/(√(2eV) × √(2V/u)) = 2√2:1.
Q.9Hard
The phenomenon where the inductance of a coil changes with the current flowing through it due to non-linear magnetic properties of the core is called:
Answer: B
When the magnetic core saturates, further increase in current produces minimal increase in magnetic flux, causing inductance to decrease. This is the saturation effect in magnetic cores.
Q.10Hard
A magnetic field B is applied perpendicular to a conductor carrying current I. The Hall coefficient is related to:
Answer: A
Hall coefficient R_H = 1/(ne), where n is charge carrier density and e is elementary charge
Q.11Hard
A charged particle enters a uniform magnetic field region at an angle θ to the field direction. Its trajectory is:
Answer: A
Velocity component parallel to B is unaffected; perpendicular component causes circular motion, resulting in helical trajectory
Q.12Hard
A superconductor exhibits the Meissner effect, which means:
Answer: A
Meissner effect: superconductor actively expels magnetic flux from its interior (B = 0), not just zero resistance
Q.13Hard
A toroidal coil has N turns and inner radius r₁, outer radius r₂. The self-inductance is approximately:
Answer: A
For a toroidal coil: L = (μ₀N²h/(2π)) × ln(r₂/r₁), where h is the height of the toroid
Q.14Hard
In a cyclotron, the time period of revolution of a particle is independent of its energy because:
Answer: A
T = 2πm/(qB), independent of v and r. As energy increases, velocity and radius increase proportionally, keeping period constant
Q.15Hard
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.16Hard
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.17Hard
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.18Hard
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)
