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A square loop of wire of edge a carries ...

A square loop of wire of edge a carries a current `i`.
Magnetic induction `B` for a point on the axis of the loop at a distance `x=a/sqrt2` from its centre is `(Nmu_(0)i)/(3pia)`,then find value of `N B=(4mu_(0)ia^(2))/(pi(4x^(2)+a^(2))(4x^(2)+2a^(2))^(1//2))`

Text Solution

Verified by Experts

The correct Answer is:
`N=2`

(a)`B_(R)=4 Bsin theta`
`=4xx(mu_(0)i)/(4pir sin alpha)[cos alpha+cos alpha]xx sin theta`
`(2mu_(0)i)/(pi[x^(2)+a^(2)/4]^(1/2))xx(a/2)/([x^(2)+a^(2)/2]^(1/2))xx(a/2)/([x^(2)+a^(2)/2]^(1/2))`
`(4mu_(0)a^(2)i)/(pi[4x^(2)+a^(2)][4x^(2)+2a^(2)]^(1/2))`
(b)`x=0`
`B_(R)=(4mu_(0)a^(2)i)/(pi (a^(2))(sqrt2a))`
`=2sqrt2(mu_(0)i)/(pia)`
(c) `x gt gt a`
`B=(4mu_(0)a^(2)i)/(pi4x^(2)(2-x))=(mu_(0)a^(2)i)/(2pix^(3))=(mu_(0)M)/(2pix^(3))`

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