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The figure shows a conducting loop ABCDA...

The figure shows a conducting loop ABCDA placed in a uniform magnetic field perpendicular to its plane. The part ABC is the `(3//4)^(th)` portion of the square of side lengthl. The part ADC is a circular are. Of radius R. The points A and C are connected to a battery which supply a current I to the circuit. The magnetic force on the loop due to the field B is

A

zero

B

Bil

C

2BIR

D

`(BI//R)/(l+R)`

Text Solution

Verified by Experts

The correct Answer is:
B
Introducing two equal and opposite current `I_(1) and also I_(2)` between A & C
Force on ABCA closed loop zero
Force on ADCA closed loop zero
Force on extra `I_(1) & I_(2)`
`F=(I_(1)+L_(2)) lB=I//B`
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