A straight segment `OC`(of length L meter) of a circuit carrying a current `I amp` is placed along the ` x- axis` ( fig.). Two infinetely long straight wires ` A and B` , each extending from ` z = -oo to +oo`, are fixed at ` y = -ameter and y = +a meter` respectively, as shown in the figure.
If the wires ` A and B` each carry a current `I amp` into the plane of the paper, obtain the expression for the force acting on the segment ` OC`. What will be the force on `OC` if the current in the wire `B` is reversed?

Options

A. `((mu_(0)I^(2))/(pi)) 1n ((L^(2)+a^(2))/(a^(2)))(-hat(k))`
B. `((mu_(0)I^(2))/(2pi)) 1n ((L^(2)+a^(2))/(a^(2)))(-hat(k))`
C. zero
D. None of these

Answer

C

Explanation

The magnetic field produced on different opints on OC will be differed. Let us consider an arbitray point P on OC which is at a distance x from the origin. Let the magnetic field due to currents in A and B at P be `B_(1)` and `B_(2)` respetively, both being in the X-Y plane

Let `/_BPO=/_APO=theta`
`|vec(B)_(1)|=(mu_(0))/(4pi)(2l)/(sqrt(a^(2)+x^(2))=|vec(B)_(2)|`
On resolving `B_(1)` and `B_(2)` we get that the `sin theta` components cencel out and the `cos theta` component add up. Therefore, the total magnetic field at P is
`:. B=2B_(1) cos theta = (2 mu_(0))/(4pi) (2l)/(sqrt(a^(2)+x^(2))) xx (x)/(sqrt(a^(2)+x^(2)))`
`=(mu_0)/(4pi) (4Ix)/(a^(2)+x^(2))`
(towards Y direction) Let us consider a small portion of wire OC at P of length dx. The small amount of force acting on the small portion `vec(dF)=I(vec(vx) xx vec(B)) :. dF= I dx B sin 90^(@)`
`implies dF=Idx (mu_0)/(4pi) xx (al)/(a^(2)+x^(2))`
The total force
`F=(mu_0)/(4pi) 4I^(2) int (xdx)/(a^(2)+x^(2)) F=(mu_(0))/(4 pi) xx 4 I^(2)[1/2 log_(e)(a^(2)+x^(2))]_(0)^(L)`
`implies F=(mu_0)/(4pi) xx 2I^(2) [ log_(e) ((a^(2)+x^(2))/(a^2))]`
To find the direction of force, we can use Fleming's left hand rule. It proves that the direction of `vec(F)` is toward Z direction. When the current in wire B is reversed, the resulant magnetic field at any arvitrary point P on OC will be in the X direction therefore, force acting will be zero `(F=IlB sin theta` and `theta=180^(@))`