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The straight wire AB carries a current I...

The straight wire `AB` carries a current `I`. The ends of the wire subtend angles `theta_1` and `theta_2` at the point `P` as shown in figure. The magnetic field at the point `P` is

A

`(mu_(0)I)/(4pia)(sin theta_(1) -sin theta_(2))`

B

`(mu_(0)I)/(4 pi a)(sin theta_(1)+sin theta_(2))`

C

`(mu_(0)I)/(4pia)(cos theta_(1)-cos theta_(2))`

D

`(mu_(I))/(4pia)(cos theta_(1)+cos theta_(2))`

Text Solution

Verified by Experts

The correct Answer is:
A
`B=(mu_(0)I)/(4pia) [ sin theta_(1)+sin (-theta_(2))]`
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