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  • Shown in figwe, is a very long semi-cylindrical conducting shell of radius and carrying a current <img alt="i." src="https://

Shown in figwe, is a very long semi-cylindrical conducting shell of radius R and carrying a current i. An infinitely long straight current carrying cunducter is lying along the axis of the seni-cylinder. If the current flowing through the straight current carrying cunducter is lying in straight wire be i_o, then the force pere unit length on the conducting wire is -

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

None


Answers (1)

best_answer

Look at the figure, consider an elemental current di, The net magnetic force on the cunducting wire per unit length - 

\Rightarrow f=\int_0^{\pi / 2} 2 d f \cos \theta \Rightarrow f=\int_0^{\pi / 2} 2\left[\frac{u_0(d i) I_0}{2 \pi R}\right] \cos \theta

\Rightarrow f=\frac{ \mu _0 i_0}{\pi R} \int_0^{\pi / 2} d i \cos \theta

    when d i=\frac{i}{\pi R} \times R d \theta=\frac{I d \theta}{\pi}

\Rightarrow F=\frac{\mu_0 i_0}{\pi R} \int_0^{\pi / 2} \frac{(i d \theta) \cos \theta}{\pi}

\Rightarrow f=\frac{\mu_0 i_0 i}{\pi^2 R} \int_{0}^{\pi / 2} \cos \theta d \theta=\frac{\mu_0 i_0 i}{\pi^2 R}

Posted by

Rakesh

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