Physics · Moving Charges and Magnetic Field

JEE Main 2024 — 4 April, Shift 1 — Question 51

The magnetic field existing in a region is given by B⃗=0.2(1+2x)k^T\vec{B}=0.2(1+2 x) \hat{k} T. A square loop of edge 50 cm carrying 0.5 A current is placed in x−y\mathrm{x}-\mathrm{y} plane with its edges parallel to the x−yx-y axes, as shown in figure. The magnitude of the net magnetic force experienced by the loop is \qquad mN .

Question figure

Answer: 50

Numerical answer — enter this value.

Step-by-step solution

Force on segment parallel to x -axis will cancel each other. Hence Fnet F_{\text {net }} will be due to portion parallel to y -axis.

F=0.5×0.5×6×0.2−0.5×0.5×0.2×5\mathrm{F}=0.5 \times 0.5 \times 6 \times 0.2-0.5 \times 0.5 \times 0.2 \times 5

=0.5×0.5×0.2=0.5 \times 0.5 \times 0.2

=0.25×0.2=0.25 \times 0.2

=50×10−3 N=50 \times 10^{-3} \mathrm{~N}

=50mN=50 \mathrm{mN}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Physics
Chapter
Moving Charges and Magnetic Field
Topic
Force and Torque on Wires and Loops, Magnetic Dipole Moment
The magnetic field existing in a region is given by vec B =0.2(1+2 x)… | JEE Main 2024 PYQ with Solution · DhiX AI