A semi circular current carrying wire having radius R is placed i… - Vidyadip

MCQ

A semi circular current carrying wire having radius $R$ is placed in $x-y$ plane with its centre at origin $‘O’$. There is non-uniform magnetic $\vec B = \frac{{{B_o}x}}{{2R}}\hat k$ (here $B_o$ is + $ve$ constant) is existing in the region. The magnetic force acting on semi circular wire will be along

✓
$- x-axis$
B
$+ y-axis$
C
$- y-axis$
D
$+ x-axis$

✓

Answer

Correct option: A.

$- x-axis$

a
$\mathrm{F}=\mathrm{I} \int \mathrm{R} \mathrm{d} \theta(\sin \theta \hat{\mathrm{i}}+\cos \theta \hat{\mathrm{j}}) \times \frac{\mathrm{B}_{0}}{2 \mathrm{R}}(-\mathrm{R} \cos \theta) \hat{\mathrm{k}}$

$=\frac{I B_{0} R}{2} \int_{0}^{\pi}(\sin \theta \hat{i}+\cos \theta \hat{j}) x(-\cos \theta \hat{k}) d \theta$

$=\int_{0}^{\pi}\left(\sin \theta \cos \theta \hat{j}-\cos ^{2} \theta \hat{i}\right) d \theta$

$=\int_{0}^{\pi}\left(\frac{\sin 2 \theta \hat{j}}{2}-\frac{(1+\cos 2 \theta) \hat{i}}{2}\right) d \theta$

$=\left[\frac{-\cos 2 \theta}{4}\right]_{0}^{\pi} \hat{j}-\left[\frac{\theta}{2}\right]_{0}^{\pi} \hat{i}-\left[\frac{\sin 2 \theta}{4}\right]_{0}^{\pi} \hat{i}$

$=0-(\pi / 2) \hat i-0=-(\pi / 2) \hat i$

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