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JEE Physics

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

301 Q 9 Topics Take Test
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Difficulty: All Easy Medium Hard 131–140 of 301
Topics in JEE Physics
Q.131 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.132 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.

Test
Q.133 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.134 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)

Test
Q.135 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

Test
Q.136 Easy Magnetism
A solenoid has 1000 turns, length 1 m, and carries a current of 2 A. The magnetic field inside the solenoid is (μ₀ = 4π × 10⁻⁷ T·m/A):
A 8π × 10⁻⁴ T
B 2π × 10⁻³ T
C 4π × 10⁻⁴ T
D π × 10⁻³ T
Correct Answer:  A. 8π × 10⁻⁴ T
EXPLANATION

B = μ₀nI = μ₀(N/L)I = 4π × 10⁻⁷ × 1000 × 2 = 8π × 10⁻⁴ T

Test
Q.137 Easy Magnetism
The radius of circular motion of a charged particle in a magnetic field is r = mv/(qB). If the magnetic field is doubled, the radius becomes:
A r/2
B r
C 2r
D r/4
Correct Answer:  A. r/2
EXPLANATION

r ∝ 1/B. When B is doubled, r becomes r/2

Test
Q.138 Easy Magnetism
A proton and an electron, both moving with the same velocity perpendicular to a uniform magnetic field, experience forces in the ratio:
A 1:1
B 1:m_e/m_p
C m_p:m_e
D e_p:e_e
Correct Answer:  A. 1:1
EXPLANATION

Magnetic force F = qvB. Since both have same charge magnitude and velocity, forces are equal in magnitude (directions opposite due to opposite charges)

Test
Q.139 Easy Magnetism
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:
A Attractive with magnitude μ₀I₁I₂/(2πd)
B Repulsive with magnitude μ₀I₁I₂/(2πd)
C Attractive with magnitude μ₀I₁I₂/(πd)
D Zero
Correct Answer:  A. Attractive with magnitude μ₀I₁I₂/(2πd)
EXPLANATION

Parallel currents in same direction attract each other with force per unit length F/L = μ₀I₁I₂/(2πd)

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Q.140 Easy Magnetism
The magnetic field at the center of a circular loop carrying current I with radius R is given by:
A μ₀I/(2R)
B μ₀I²/(2R²)
C 2μ₀I/R
D μ₀I/(πR)
Correct Answer:  A. μ₀I/(2R)
EXPLANATION

Using Biot-Savart law for a circular loop, the magnetic field at center is B = μ₀I/(2R)

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