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Figure shows a rectangular coil near a l...

Figure shows a rectangular coil near a long wire. The mutual inductance of the combination is

A

`(mu_(0)a)/(2pi) In(1-(b)/(c))`

B

`(mu_(0)a)/(2pi) In(1+ (b)/(c))`

C

`(mu_(0)a)/(pi) In(1+ (b)/(c))`

D

`(mu_(0)a)/(sqrt(2)pi) In(1+ (b)/(c))`

Text Solution

Verified by Experts

The correct Answer is:
B

Let current `i_(1)` in the straight wire be upward. Then the magnetic field due to the straight wire has magnitude `B_(1) = mu_(0)i_(1)//2 pir` at a distance `r`. In according with right hand rule, `B_(1)` points inward to the plane of page. We consider a differential strip of thickness `dr`, area `dA_(2) = a dr`. Magnetic flux through area `dA, dphi_(B)(a dr)`.
Total flux through the loop,
`phi_(B) = int B_(1) adr = int_(c)^(c + b) (mu_(0)i_(1))/(2pir) a dr`
`= (mu_(0)i_(1)a)/(2pi) int_(c)^(c + b) (dr)/(r ) = (mu_(0)i_(1)a)/(2pi) In((c + b)/(c))`
therefore mutual inductance,
`M = (phi)/(i_(1)) = (mu_(0)a)/(2pi) In (1 + (b)/(c))`
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