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Question

A current carrying loop in the form of a right angle isosceles triangle ABC is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is F, the force on the arm AC is :


A
2F
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B
F
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C
F
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D
2F
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Solution

The correct option is B F
If the magnetic field along AB is B, the force acting on BC,

F=BIl

Now the length of the side AC is 2l.

Also the current reverses the direction in this section as compared to that in BC.

Hence, force on AC=2l(B×I)

=BIlsin45=BIl=F

Hence, (B) is the correct answer.

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