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Physics · Moving Charges and Magnetic Field

JEE Main 2024 — 9 April, Shift 1 — Question 55

A square loop of edge length 2 m carrying current of 2 A is placed with its edges parallel to the x−yx-y axis. A magnetic field is passing through the x−yx-y plane and expressed as B⃗=B0(1+4x)k^\vec{B}=B_{0}(1+4 x) \hat{k}, where B0=5 TB_{0}=5 \mathrm{~T}. The net magnetic force experienced by the loop is \qquad N .

Answer: 160

Numerical answer — enter this value.

Step-by-step solution

B(x=0)=B0,B(x=2)=9B0B(x=0) = B_0, \quad B(x=2) = 9B_0

Also, F=iℓBF = i\ell B

  ⟹  \implies F1=iℓB0&F2=9iℓB0F_1 = i\ell B_0 \quad \& \quad F_2 = 9i\ell B_0

F=F2−F1=8iℓB0=8×2×2×5F = F_2 - F_1 = 8i\ell B_0 = 8 \times 2 \times 2 \times 5

F=160F = 160 N

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
9 April, Shift 1
Subject
Physics
Chapter
Moving Charges and Magnetic Field
Topic
Force and Torque on Wires and Loops, Magnetic Dipole Moment

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