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STD 12 - 4. Moving charges and magnetism — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 4. Moving charges and magnetism4 Marks
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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