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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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