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Figure. shows a conducting loop ABCDA pl...

Figure. shows a conducting loop ABCDA placed in a uniform magnetic field (strength B) perpendicular to its plane. The part ABC is the (three-fourth) portion of the square of side length l. The part ADC is a circular arc of radius R. The points A and C are connected to a battery which supplies a current I to the circuit. The magnetic force on the loop due to the field B is

A

zero

B

`BI l`

C

`2 BI R`

D

`(BI lR)/(1+R)`

Text Solution

Verified by Experts

The correct Answer is:
B
`F=Bi_(1)l+Bi_(2)l`

`= Bl(i_(1)+i_(2))=Bil`
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