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Question

# A wire carrying a current of 3 A is bent in the form of a parabola y2=4−x as shown in figure, where x and y are in metre. The wire is placed in a uniform magnetic field →B=5^k T. The magnetic force acting on the wire is:

A
(60 ^i) N
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B
(60 ^i) N
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C
(30 ^i) N
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D
(30 ^i) N
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Solution

## The correct option is A (60 ^i) NJoining the initial and final points on the parabolic shaped wire we get the length/displacement vector as shown in figure, →L=−−→NP From the equation of parabola, y2=4−x Putting x=0 i.e for point P & N we have, ⇒y2=4 or, y=±√4 y=±2 Thus, |OP|=|ON|=2 m Hence NP=4 m, →L=(4 ^j) m Using the magnetic force relation we have: →Fm=i(→L×→B) →Fm=3[(4^j)×(5^k)] →Fm=(60 ^i) N

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