Currents of 10, A and 2, A are passed through | vedclass

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Question

(A) The magnetic field $B_A$ produced by the infinitely long wire $A$ at a distance $r$ is given by $B_A = \frac{\mu_0 I_A}{2 \pi r}$.Here,$I_A = 10\,

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$8 	imes 10^{-5}\, N$

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(A) The magnetic field $B_A$ produced by the infinitely long wire $A$ at a distance $r$ is given by $B_A = \frac{\mu_0 I_A}{2 \pi r}$.Here,$I_A = 10\, A$ and $r = 10\, cm = 0.1\, m$.So,$B_A = \frac{4\pi 	imes 10^{-7} 	imes 10}{2\pi 	imes 0.1} = 2 	imes 10^{-5}\, T$.The force $F$ on a current-carrying wire $B$ of length $L$ in a magnetic field $B_A$ is given by $F = I_B L B_A \sin(	heta)$.Since the wires are parallel,the angle $	heta = 90^\circ$ (between the current direction and the magnetic field lines),so $\sin(90^\circ) = 1$.Given $I_B = 2\, A$ and $L = 2\, m$,we have $F = 2 	imes 2 	imes (2 	imes 10^{-5}) = 8 	imes 10^{-5}\, N$.

Currents of $10\, A$ and $2\, A$ are passed through two parallel thin wires $A$ and $B$ respectively in opposite directions. Wire $A$ is infinitely long and the length of wire $B$ is $2\, m$. The force acting on the conductor $B$,which is situated at a distance of $10\, cm$ from $A$,will be:

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