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Question: Shown in the figure is a very long semicylindrical conducting shell of radius R and carrying a curre...

Shown in the figure is a very long semicylindrical conducting shell of radius R and carrying a current i.i. An infinitely long straight current carrying conductor is lying along the axis of the semicylinder. If the current flowing through the straight wire be i0i_{0} then the force on the semicylinder is

A

μ0ii0πR2\frac{\mu_{0}ii_{0}}{\pi R^{2}}

B

μ0i0iπ2R\frac{\mu_{0}i_{0}i}{\pi^{2}R}

C

μ0i02iπ2R\frac{\mu_{0}i_{0}^{2}i}{\pi^{2}R}

D

None of these

Answer

μ0i0iπ2R\frac{\mu_{0}i_{0}i}{\pi^{2}R}

Explanation

Solution

The net magnetic force on the conducting wire

== 2dFcosθ= \text{F } = \ \int_{}^{}{\text{2dFcos}\theta}

⇒ F = 2[μ0(di)i02πR]cosθ\int_{}^{}{2\left\lbrack \frac{\mu_{0}(di)i_{0}}{2\pi R} \right\rbrack\cos\theta}

⇒ F = μ0i0πRdicosθ\frac{\mu_{0}i_{0}}{\pi R}\int_{}^{}{di\cos\theta}When di=iπR x Rdθ=idθπdi = \frac{i}{\pi R}\text{ x Rd}\theta = \frac{\text{id}\theta}{\pi}

F=μ0i0πR(idθ)cosθπF = \frac{\mu_{0}i_{0}}{\pi R}\int_{}^{}\frac{(id\theta)\cos\theta}{\pi}

F=μ0ioiπ2R0π/2cosθdθ=μ0i0iπ2RF = \frac{\mu_{0}i_{o}i}{\pi^{2}R}\int_{0}^{\pi/2}{\cos\theta d\theta = \frac{\mu_{0}i_{0}i}{\pi^{2}R}}