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(a) Two long wires at a distance 2d apart carry equal, antiparallel current i. Find magnitude and direction of magnetic field at point A as shown.For what value of x, magnetic field is maximum? Also calculate maximum magnetic field. Sketch B versus x graph.

Answer»

Solution :
Magnetic field at `A` due to wire `①`
`B_(1)=(mu_(0)i)/(2pisqrt(x^(2)+d^(2))), B_(1)` will be `bot^(ar)` to `JA` at `A` as SHOWN

Magnetic field at `A` due to wire `②`
`B_(2)=(mu_(0)i)/(2pisqrt(x^(2)+d^(2))), B_(2)` will be `bot^(ar)` to `PA` at `A` as shown

`B_(A)=2B sin theta=2.(mu_(0)i)/(2pisqrt(x^(2)+d^(2))).d/(sqrt(x^(2)+d^(2)))`
`=(mu_(0)id)/(pi(x^(2)+d^(2)))`, towards left
`B_(A)` is maximum at `x=0, (B_(A))_(MAX)=(mu_(0)i)/(pid)`
Left of `O:`

The direction of magnetic field on the `x`-axis will be towards left.
Taking magnetic field `-ve` towards the negative `x`-axis

(b)
`B_(1) = B_(2) = (mu_(0)i)/(2 pi sqrt(x^(2) + d^(2))) = B`
`= (mu_(0)ix)/(pi (x^(2) + d^(2)))` , downward
For `B_(A)` to be maximum,
`(dB_(A))/(dx) = 0 RARR ((x^(2) + d^(2)). 1 - x (2x + 0))/(pi(x^(2) + d^(2))^(2)) = 0`
`x^(2) = d^(2) rArr x = -+ d`
`(B_(A))_("max") = (mu_(0)i)/(2 pi d)`
`x rarr OO, B rarr 0, x rarr - oo, B rarr 0`
Direction of magnetic field left of O:

Left of O, magnetic field will be upward, taking upward direction positive.


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