Start free

Physics · Moving Charges and Magnetic Field

JEE Main 2024 — 31 January, Shift 2 — Question 30

A uniform magnetic field of 2×10−3 T2 \times 10^{-3} \mathrm{~T} acts along positive Y-direction. A rectangular loop of sides 20 cm and 10 cm with current of 5 A is Y−Z\mathrm{Y}-\mathrm{Z} plane. The current is in anticlockwise sense with reference to negative X axis. Magnitude and direction of the torque is :

  1. Option A:

    2×10−4 N−m2 \times 10^{-4} \mathrm{~N}-\mathrm{m} along positive Z -direction

  2. Option B:

    2×10−4 N−m2 \times 10^{-4} \mathrm{~N}-\mathrm{m} along negative Z -direction

    Correct
  3. Option C:

    2×10−4 N−m2 \times 10^{-4} \mathrm{~N}-\mathrm{m} along positive X -direction

  4. Option D:

    2×10−4 N−m2 \times 10^{-4} \mathrm{~N}-\mathrm{m} along positive Y -direction

Answer: B

Step-by-step solution

M→=iA→\overrightarrow{\mathrm{M}}=\mathrm{i} \overrightarrow{\mathrm{A}}

=5×(0.2)×(0.1)(−i^)=5 \times(0.2) \times(0.1)(-\hat{\mathrm{i}}) =0.1(−i^)=0.1(-\hat{\mathrm{i}})

τ⃗=M→×B→=0.1(−i^)×(2×10−3)(j^)\vec{\tau}=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}=0.1(-\hat{\mathrm{i}}) \times\left(2 \times 10^{-3}\right)(\hat{\mathrm{j}})

=2×10−4(−k^)N−m=2 \times 10^{-4}(-\hat{\mathrm{k}}) \mathrm{N}-\mathrm{m}

Solution figure

Answer key and solution verified before publishing.

Practise Moving Charges and Magnetic Field

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2024
Paper
31 January, Shift 2
Subject
Physics
Chapter
Moving Charges and Magnetic Field
Topic
Force and Torque on Wires and Loops, Magnetic Dipole Moment

More Moving Charges and Magnetic Field questions from this paper

A uniform magnetic field of 2 × 10 -3 T acts along positive… | JEE Main 2024 PYQ with Solution · DhiX AI