The adjoining figure shows a very long semi-c | vedclass

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(A) Consider a small element of the semi-cylindrical shell subtending an angle $d	heta$ at the axis. The current in this element is $di = \frac{i}{\p

Vedclass · Accepted Answer

$\frac{\mu_0 i i_0}{\pi^2 R}$

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(A) Consider a small element of the semi-cylindrical shell subtending an angle $d	heta$ at the axis. The current in this element is $di = \frac{i}{\pi R} \cdot R d	heta = \frac{i d	heta}{\pi}$.The magnetic field produced by the straight wire at the shell is $B = \frac{\mu_0 i_0}{2\pi R}$.The force on the element of length $L$ is $dF = B(di)L$. For unit length $(L=1)$,$dF = \frac{\mu_0 i_0}{2\pi R} \cdot \frac{i d	heta}{\pi} = \frac{\mu_0 i i_0}{2\pi^2 R} d	heta$.Due to symmetry,the horizontal components of the force cancel out,and the vertical components add up. The vertical component is $dF_y = dF \cos	heta$.Integrating from $-\pi/2$ to $\pi/2$ is equivalent to $2 \int_0^{\pi/2} dF \cos	heta$.$F = 2 \int_0^{\pi/2} \frac{\mu_0 i i_0}{2\pi^2 R} \cos	heta d	heta = \frac{\mu_0 i i_0}{\pi^2 R} [\sin	heta]_0^{\pi/2} = \frac{\mu_0 i i_0}{\pi^2 R}$.

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