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A long straight wire, carrying current I...

A long straight wire, carrying current I, is bent at its midpoint to form an angle of `45^@`. Find the induction of magnetic field at point P, distant R from the point of bending (as shown in)

A

`((sqrt(2)-1)mu_(0)l)/(4piR)`

B

`((sqrt(2)+1)mu_(0)l)/(4sqrt(2)piR)`

C

`(sqrt(2-1)mu_(0)l)/(4sqrt(2)piR)`

D

`((sqrt(2)+1)mu_(0)l)/(4sqrt(2)piR)`

Text Solution

Verified by Experts

The correct Answer is:
A
`therefore` B (at P) `=(mu_(0)I)/(4pid) (costheta_(1)-costheta_(2))`

In given case, `d=Rsin45^(@)=R/sqrt(2)`
`theta_(1) = 135^(@), theta_(2)=180^(@)`
`therefore` B(at P) `=(mu_(0)I)/(4pi(R/sqrt(2)))[cos135^(@)-cos180^(@)]`
`=(mu_(0)I)/(4piR)sqrt(2)(-1/sqrt(2)-(-1))=(mu_(0)I)/(4piR)sqrt(2)((sqrt(2)-1)/sqrt(2))`
or B (at P) `=(mu_(0)I)/(4piR)(sqrt(2)-1)T`
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