A point charge is brought from infinity to a point in an electric field. Work done depends on:

The correct answer is C: Initial and final positions only. Electric force is conservative; work depends only on initial and final positions (potentials), not on the path taken. W = q(V_initial - V_final).

A lens has power +5 D. A second lens of power -2 D is placed in contact with it. The equivalent focal length of the combination is:

The correct answer is A: 0.33 m. Power combination: P = P₁ + P₂ = 5 + (-2) = 3 D. Focal length f = 1/P = 1/3 = 0.33 m

A heat engine operates between 600 K and 300 K reservoirs. It absorbs 5000 J from hot reservoir. For a Carnot engine operating between same temperatures, maximum work output would be:

The correct answer is B: 2500 J. Carnot efficiency = 1 - 300/600 = 0.5; W_max = η × Q_h = 0.5 × 5000 = 2500 J

Two parallel wires carrying currents I₁ and I₂ in the same direction are separated by distance d. The force per unit length between them is:

The correct answer is A: Attractive with magnitude μ₀I₁I₂/(2πd). Parallel currents in same direction attract each other with force per unit length F/L = μ₀I₁I₂/(2πd)

The specific contact resistance of a metal-semiconductor junction is proportional to:

The correct answer is B: exp(-φB/kT). Contact resistance ρc ∝ exp(φB/kT)/Nc where φB is barrier height. Higher barrier leads to exponentially higher resistance following Thermionic emission theory.

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Magnetism

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

37 Q 9 Topics Take Mock Test

Difficulty: All Easy Medium Hard 21–30 of 37

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.21 Easy Magnetism

Two parallel wires carrying currents I₁ and I₂ in the same direction, separated by distance d. The force per unit length between them is:

A F/L = μ₀I₁I₂/(2πd²)

B F/L = μ₀I₁I₂/(πd)

C F/L = μ₀I₁I₂/(2πd)

D F/L = 2μ₀I₁I₂/(πd)

Correct Answer: C. F/L = μ₀I₁I₂/(2πd)

EXPLANATION

Magnetic force per unit length between parallel wires: F/L = μ₀I₁I₂/(2πd), attractive for same direction currents

Test

Q.22 Easy Magnetism

A proton moving with velocity v perpendicular to a magnetic field B follows a circular path. The radius of this path is:

A r = mv/(qB)

B r = qB/(mv)

C r = qvB/m

D r = v/(qB)

Correct Answer: A. r = mv/(qB)

EXPLANATION

For circular motion in magnetic field: qvB = mv²/r, therefore r = mv/(qB)

Test

Q.23 Easy Magnetism

The magnetic field intensity at a distance r from a long straight current-carrying wire is:

A B = μ₀I/(4πr)

B B = μ₀I/(2πr)

C B = μ₀I/r

D B = μ₀I/(2r)

Correct Answer: B. B = μ₀I/(2πr)

EXPLANATION

Using Ampere's circuital law for an infinite straight wire: B = μ₀I/(2πr), where μ₀ = 4π × 10⁻⁷ T·m/A

Test

Q.24 Easy Magnetism

A magnetic dipole of moment M is placed in a uniform magnetic field B. The torque experienced by the dipole when it makes an angle θ with the field is:

A τ = MB sin θ

B τ = MB cos θ

C τ = MB tan θ

D τ = MB/sin θ

Correct Answer: A. τ = MB sin θ

EXPLANATION

Torque on a magnetic dipole: τ = M × B = MB sin θ, where θ is the angle between M and B.

Test

Q.25 Easy Magnetism

The magnetic field due to a solenoid inside and outside differs significantly. Which statement is correct?

A Field inside is uniform and strong; outside is zero

B Field inside is zero; outside is strong

C Field is equal inside and outside

D Field inside is non-uniform and weak

Correct Answer: A. Field inside is uniform and strong; outside is zero

EXPLANATION

An ideal solenoid produces a uniform, strong magnetic field inside (B = μ₀nI) and negligible field outside. This is due to cancellation of fields from adjacent turns outside.

Test

Q.26 Easy Magnetism

According to Lenz's law, the induced EMF in a coil opposes the change in magnetic flux. If the flux through a coil increases at a rate of 0.1 Wb/s and the coil has 100 turns, the induced EMF is:

A 10 V

B 1 V

C 0.1 V

D 100 V

Correct Answer: A. 10 V

EXPLANATION

Induced EMF ε = -N × dΦ/dt = -100 × 0.1 = -10 V. The magnitude is 10 V.

Test

Q.27 Easy Magnetism

The magnetic flux through a surface is Φ = ∫B·dA. For a uniform field B perpendicular to a rectangular surface of area 2 m², with B = 0.5 T, the flux is:

A 0 Wb

B 0.5 Wb

C 1 Wb

D 2 Wb

Correct Answer: C. 1 Wb

EXPLANATION

Φ = BA cos θ. Since B is perpendicular to the surface, θ = 0° and cos θ = 1. Φ = 0.5 × 2 = 1 Wb.

Test

Q.28 Easy Magnetism

A charged particle with charge q and mass m enters a magnetic field region with velocity v perpendicular to B. The radius of circular motion is given by:

A r = mv/(qB)

B r = qB/(mv)

C r = v/(qBm)

D r = qv/(mB)

Correct Answer: A. r = mv/(qB)

EXPLANATION

From qvB = mv²/r, we get r = mv/(qB). This is the radius of curvature for a charged particle in a perpendicular magnetic field.

Test

Q.29 Easy Magnetism

An electron moves in a circular path in a perpendicular magnetic field. Which quantity remains constant during its motion?

A Kinetic energy

B Velocity

C Speed

D Acceleration

Correct Answer: A. Kinetic energy

EXPLANATION

Magnetic force is always perpendicular to velocity, so it does no work. Kinetic energy (½mv²) remains constant, but direction changes, so velocity changes while speed remains constant.

Test

Q.30 Easy Magnetism

The SI unit of magnetic field intensity is Tesla. One Tesla is equivalent to which of the following?

A kg/(A·s²)

B kg·m/(A·s²)

C kg/(A²·s)

D kg·s/(A·m)

Correct Answer: A. kg/(A·s²)

EXPLANATION

1 Tesla = 1 Weber/m² = 1 kg/(A·s²). This is the correct dimensional formula for magnetic field intensity.

Test

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