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JEE Physics
Magnetism

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

37 Q 9 Topics Take Mock Test
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Difficulty: All Easy Medium Hard 1–10 of 37
Topics in JEE Physics
Q.1 Easy Magnetism
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:
A eBr
B eBr/2
C 2eBr
D eBr²
Correct Answer:  A. eBr
EXPLANATION

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).

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Q.2 Easy Magnetism
The magnetic moment of a current loop is defined as:
A M = IA
B M = I/A
C M = IA²
D M = I√A
Correct Answer:  A. M = IA
EXPLANATION

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.

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Q.3 Easy Magnetism
A solenoid has n turns per unit length and carries current I. The magnetic field inside the solenoid is:
A μ₀nI
B μ₀nI/2
C μ₀I/n
D μ₀I/(2πn)
Correct Answer:  A. μ₀nI
EXPLANATION

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.

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Q.4 Easy Magnetism
A charged particle enters a uniform magnetic field with velocity perpendicular to the field. The particle will move in:
A A straight line
B A circular path
C A parabolic path
D An elliptical path
Correct Answer:  B. A circular path
EXPLANATION

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).

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Q.5 Easy Magnetism
Two parallel wires carry currents I₁ and I₂ in the same direction, separated by distance d. The force per unit length between them is:
A μ₀I₁I₂/(2πd) (repulsive)
B μ₀I₁I₂/(2πd) (attractive)
C μ₀(I₁+I₂)/(2πd)
D μ₀I₁I₂/(πd²)
Correct Answer:  B. μ₀I₁I₂/(2πd) (attractive)
EXPLANATION

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.

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Q.6 Easy Magnetism
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:
A BIL
B BIL sinθ
C BIL cosθ
D BIL tanθ
Correct Answer:  B. BIL sinθ
EXPLANATION

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.

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Q.7 Easy Magnetism
Which of the following is NOT a property of magnetic field lines?
A They form closed loops
B They never intersect each other
C They are denser where the field is stronger
D They can exist in isolation like electric field lines
Correct Answer:  D. They can exist in isolation like electric field lines
EXPLANATION

Magnetic field lines always form closed loops and cannot exist in isolation. Unlike electric field lines which start from positive charges and end at negative charges, magnetic field lines have no beginning or end.

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Q.8 Easy Magnetism
A magnetic field B is perpendicular to a plane containing a circular loop of radius r. If the magnetic field increases uniformly from 0 to B₀ in time t, what is the magnitude of induced EMF in the loop?
A πr²B₀/t
B 2πrB₀/t
C πrB₀/t
D r²B₀/2πt
Correct Answer:  A. πr²B₀/t
EXPLANATION

By Faraday's law, induced EMF = -dΦ/dt. Magnetic flux Φ = BA = πr²B. As B changes from 0 to B₀ in time t, EMF = πr²(B₀-0)/t = πr²B₀/t

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Q.9 Easy Magnetism
A conducting rod of length L moves with velocity v perpendicular to a magnetic field B. The induced EMF across the rod is:
A BLv
B BL/v
C BL²v
D Bv/L
Correct Answer:  A. BLv
EXPLANATION

Motional EMF = BLv (where L is perpendicular to both B and v)

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Q.10 Easy Magnetism
A galvanometer can be converted into an ammeter by connecting a:
A Low resistance in parallel (shunt)
B High resistance in series (multiplier)
C Capacitor in series
D Inductor in parallel
Correct Answer:  A. Low resistance in parallel (shunt)
EXPLANATION

A shunt (low resistance in parallel) diverts excess current, protecting the galvanometer coil

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