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JEE Physics

Physics questions for JEE Main — Mechanics, Electrostatics, Optics, Modern Physics.

165 Q 9 Topics Take Test
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Difficulty: All Easy Medium Hard 91–100 of 165
Topics in JEE Physics
Q.91 Hard Magnetism
A superconductor exhibits the Meissner effect, which means:
A Magnetic field lines are expelled from the interior
B Magnetic field is trapped inside permanently
C No current flows through it
D Resistance becomes infinite
Correct Answer:  A. Magnetic field lines are expelled from the interior
EXPLANATION

Meissner effect: superconductor actively expels magnetic flux from its interior (B = 0), not just zero resistance

Test
Q.92 Hard Magnetism
A charged particle enters a uniform magnetic field region at an angle θ to the field direction. Its trajectory is:
A Helical path with axis parallel to B
B Circular path in the plane perpendicular to B
C Straight line along B
D Parabolic path
Correct Answer:  A. Helical path with axis parallel to B
EXPLANATION

Velocity component parallel to B is unaffected; perpendicular component causes circular motion, resulting in helical trajectory

Test
Q.93 Hard Magnetism
A magnetic field B is applied perpendicular to a conductor carrying current I. The Hall coefficient is related to:
A Charge carrier density and elementary charge
B Only resistance of conductor
C Only magnetic field strength
D Temperature only
Correct Answer:  A. Charge carrier density and elementary charge
EXPLANATION

Hall coefficient R_H = 1/(ne), where n is charge carrier density and e is elementary charge

Test
Q.94 Hard Magnetism
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:
A Self-inductance
B Saturation effect
C Mutual inductance
D Hysteresis loss
Correct Answer:  B. Saturation effect
EXPLANATION

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.

Test
Q.95 Hard Magnetism
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:
A 2:1
B √2:1
C 2√2:1
D 1:2
Correct Answer:  C. 2√2:1
EXPLANATION

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.

Test
Q.96 Hard Magnetism
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:
A U = μ₀N²I²A/(2πR)
B U = μ₀N²I²/(2πRA)
C U = μ₀NI²A/(πR)
D U = μ₀N²I/(4πR)
Correct Answer:  A. U = μ₀N²I²A/(2πR)
EXPLANATION

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).

Test
Q.97 Hard Magnetism
The magnetic field at the center of a circular arc of radius R subtending angle θ at the center and carrying current I is:
A B = μ₀Iθ/(4πR)
B B = μ₀I/(2R)
C B = μ₀I/(4πR)
D B = μ₀Iθ/(2R)
Correct Answer:  A. B = μ₀Iθ/(4πR)
EXPLANATION

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.

Test
Q.98 Hard Magnetism
The Hall effect in semiconductors is used to determine:
A Only the sign of charge carriers
B Only the number density of charge carriers
C Both sign and number density of charge carriers
D Only the mobility of charge carriers
Correct Answer:  C. Both sign and number density of charge carriers
EXPLANATION

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.

Test
Q.99 Hard Magnetism
Two identical coils are placed coaxially with separation much larger than their radius. Their mutual inductance is:
A M = μ₀N₁N₂A/d
B M = μ₀N₁N₂A/d²
C M = μ₀N₁N₂/d
D Cannot be determined without more info
Correct Answer:  B. M = μ₀N₁N₂A/d²
EXPLANATION

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.

Test
Q.100 Hard Magnetism
The magnetic field inside a toroid with N turns, major radius R, and carrying current I is:
A B = μ₀NI/2πR
B B = μ₀NI/R
C B = μ₀I/2πR
D B varies with distance from center
Correct Answer:  D. B varies with distance from center
EXPLANATION

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.

Test
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