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Q.

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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a

3IB0R

b

2IB0R

c

10IB0R

d

5IB0R

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 F1=iB0(2R) Force experienced by BC is, F2=i×2B0×2RResultant magnetic force acting on the wire is,F=F12+F22=10IB0R
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