A semicircular wire of radius R, carrying current i, is placed in a magnetic field as shown in figure. On left side of X'X, magnetic field strength is B₀, and on right side of X'X,magnetic field strength is 2B₀. Both fields are directed inside the page.The magnetic force experienced by the wire would be

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$3 I B_{0} R$

$2 I B_{0} R$

$\sqrt{10} I B_{0} R$

$\sqrt{5} I B_{0} R$

answer is C.

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Detailed Solution

The part AUB of the wire can be replaced by straight wire AB and BWC can be replaced by BC.

Force experienced by AB is $F_{1} = i B_{0} (\sqrt{2} R)$

$Force experienced by B C is, F_{2} = i \times 2 B_{0} \times \sqrt{2} R$

Resultant magnetic force acting on the wire is,

$F = \sqrt{F_{1}^{2} + F_{2}^{2}} = \sqrt{10} I B_{0} R$

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