Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A straight segment OC (of length L) of a circuit carrying a current I is placed along the .v-axis . Two infinitely long straight wires A and B, each extending from z = - $-\infty to +\infty , $ are fixed at y= - a and y = + a respectively, as shown in the figure. If the wires A and B each carry a current 1 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?Physics Question Image

IIT JEEIIT JEE 1992Moving Charges and Magnetism

Solution:

(a) Let us assume a segment of wire OC at a point P, a
distancex from the centre of length dx as shown in figure.
Magnetic field at P due to current in wires A and B will be
in the directions perpendicular to AP and BP respectively
as shown.
$| B |=\frac{\mu_0}{2\pi}\frac{I}{AP}$
Therefore, net magnetic force at P will be along negative
v-axis as shown
$\, \, \, \, \, \, \, \, \, \, \, \, B_{net}=2| B | cos\theta$
$\, \, \, \, \, \, \, \, \, \, \, \, =2\Big(\frac{\mu_0}{2\pi}\Big)\frac{I}{AP}\Big(\frac{x}{AP}\Big)$
$\, \, \, \, \, \, \, \, \, \, \, \, B_{net}=\Big(\frac{\mu_0}{\pi}\Big)\frac{1.x}{(AP)^2}$
$\, \, \, \, \, \, \, \, \, \, \, \, B_{net}=\frac{\mu_0}{\pi}.\frac{1x}{(a^2+x^2)}$
Therefore, force on this element will be
$dF=1\Bigg\{\frac{\mu_0}{\pi}\frac{1x}{a^2+x^2}\Bigg\}dx \, \, \, \, \, \, \, \, \, \, \, $(in negative z-direction)
$\therefore $ Total force on the wire will be
$\, \, \, \, \, \, \, \, F=\textstyle \int_{x=0}^{x=L}$ dF=\frac{\mu_0 I^2}{\pi}$\textstyle \int_{0}^{L}\frac{xdx}{x^2+a^2}$
$=\frac{\mu_0 I^2}{2\pi}In\Bigg(\frac{L^2+a^2}{a^2}\Bigg)\, \, \, \, \, \, \, $(in negative z-axis)
Hence, $\, \, \, \, \, F=-\frac{\mu_0 I^2}{2\pi}In\Bigg(\frac{L^2+a^2}{a^2}\Bigg)\widehat{k}$
(b) When direction of current in B is reversed net magnetic
field is along the current. Hence, force is zero.

Solution Image Solution Image