A square loop of edge length 2 m carrying current of 2 A is placed with its edges parallel to the x-y axis. A magnetic field is passing through the x-y plane and expressed as B =B0(1+4 x) k, where B0=5 T. The net magnetic force experienced by the loop is _______ N.

JEE Mains · Physics · STD 12 - Moving charges and magnetism

A square loop of edge length \(2 \mathrm{~m}\) carrying current of \(2 \mathrm{~A}\) is placed with its edges parallel to the \(\mathrm{x}-\mathrm{y}\) axis. A magnetic field is passing through the \(x-y\) plane and expressed as \(\vec{B}=B_0(1+4 x) \hat{k}\), where \(\mathrm{B}_0=5 \mathrm{~T}\). The net magnetic force experienced by the loop is _______ \(\mathrm{N}\).

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Verified Solution

Correct Answer

(B) \(160\)

Detailed explanation

\(\mathrm{B}(\mathrm{x}=0)=\mathrm{B}_0, \quad \mathrm{~B}(\mathrm{x}=2)=9 \mathrm{~B}_0\)

\(\text { Also, } \mathrm{F}=\mathrm{i} \ell \mathrm{B}\)

\(\Rightarrow \mathrm{F}_1=\mathrm{i} \ell \mathrm{B}_0 \& \mathrm{~F}_2=9 \mathrm{i} \ell \mathrm{B}_0\)

\(\mathrm{~F}=\mathrm{F}_2-\mathrm{F}_1=8 \mathrm{i} \ell \mathrm{B}_0=8 \times 2 \times 2 \times 5\)

\(\mathrm{~F}=160 \mathrm{~N}\)

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