Two identical metal spheres with charges +Q and -Q are brought in contact and then separated. What is the final charge on each sphere?

The correct answer is B: 0 and 0. When identical conducting spheres touch, charge distributes equally. Total charge = Q - Q = 0. Each sphere gets 0 charge.

In an n-channel JFET (Junction Field Effect Transistor), when the gate-source voltage (Vgs) becomes more negative, what happens to the channel conductance?

The correct answer is B: Conductance decreases due to depletion region expansion. More negative Vgs increases the reverse bias on the gate junction, expanding the depletion region and narrowing the conducting channel, thus decreasing conductance.

A block slides down a frictionless incline of height 5 m. What is its speed at the bottom? (g = 10 m/s²)

The correct answer is B: 10 m/s. Using energy conservation: mgh = ½mv². v = √(2gh) = √(2×10×5) = √100 = 10 m/s

A battery of EMF E and internal resistance r is connected to an external resistance R. The terminal voltage is:

The correct answer is D: Both B and C. Terminal voltage V = E - Ir. Since I = E/(R+r), we get V = E - Er/(R+r) = ER/(R+r). Both expressions are equivalent

Two magnets are placed with their north poles facing each other. The force between them varies with distance r as:

The correct answer is D: F ∝ 1/r⁴. 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.

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JEE Physics
Magnetism

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

19 Q 9 Topics Take Mock Test

Difficulty: All Easy Medium Hard 1–10 of 19

Topics in JEE Physics

All Mechanics 100 Thermodynamics 100 Electrostatics 100 Current Electricity 100 Magnetism 100 Optics 100 Modern Physics 100 Semiconductors 100 Waves 100

Q.1 Hard Magnetism

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:

A V²NL/(2R)

B V²μ₀N²A/(2RL)

C V²μ₀N²A/(2R²L)

D V²L/(2Rμ₀N²A)

Correct Answer: C. V²μ₀N²A/(2R²L)

EXPLANATION

Current I = V/R. Self-inductance L = μ₀N²A/L. Magnetic energy = LI²/2 = (μ₀N²A/L)·(V²/R²)/2 = V²μ₀N²A/(2R²L)

Test

Q.2 Hard Magnetism

A charged particle enters a region with perpendicular electric and magnetic fields with velocity v. For the particle to pass undeflected, the condition is:

A E = vB

B E = B/v

C E = v/B

D E = B²/v

Correct Answer: A. E = vB

EXPLANATION

For undeflected motion, electric force equals magnetic force: qE = qvB, which gives E = vB. This is the principle of a velocity selector.

Test

Q.3 Hard Magnetism

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:

A Zero

B B₀kbIa

C B₀kIab

D 2B₀kIa

Correct Answer: C. B₀kIab

EXPLANATION

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).

Test

Q.4 Hard Magnetism

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:

A 1:1

B 1:2

C 2:1

D 1:4

Correct Answer: C. 2:1

EXPLANATION

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) = √(1/4)·√2 = √(1/2)·√2 = 2/1

Test

Q.5 Hard Magnetism

In a cyclotron, the time period of revolution of a particle is independent of its energy because:

A The centripetal force equals magnetic force, and they both change proportionally with velocity

B The magnetic field is constant

C The particle mass is constant

D The radius increases with energy in a specific way

Correct Answer: A. The centripetal force equals magnetic force, and they both change proportionally with velocity

EXPLANATION

T = 2πm/(qB), independent of v and r. As energy increases, velocity and radius increase proportionally, keeping period constant

Test

Q.6 Hard Magnetism

A toroidal coil has N turns and inner radius r₁, outer radius r₂. The self-inductance is approximately:

A L = (μ₀N²h/(2π)) × ln(r₂/r₁)

B L = μ₀N²πr²/h

C L = μ₀N²/r

D L = μ₀N²h/r₁r₂

Correct Answer: A. L = (μ₀N²h/(2π)) × ln(r₂/r₁)

EXPLANATION

For a toroidal coil: L = (μ₀N²h/(2π)) × ln(r₂/r₁), where h is the height of the toroid

Test

Q.7 Hard Magnetism

A conducting rod AB of length 1 m moves with velocity 5 m/s perpendicular to a magnetic field of 2 T. If the rod makes an angle of 60° with the direction of motion, the induced EMF is:

A 5√3 V

B 10 V

C 5 V

D 2.5 V

Correct Answer: C. 5 V

EXPLANATION

Effective length perpendicular to motion = 1 × sin(60°) = √3/2... Actually EMF = BLv cos(30°) = 2 × 1 × 5 × cos(30°) = 5√3 V. Wait, reconsidering: if angle is 60° with motion, then EMF = BLv = 2 × 1 × 5 = 10 V when fully perpendicular. The answer involves geometry nuance. Standard formula gives 10 V for full perpendicularity.

Test

Q.8 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.9 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.10 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

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