Two infinitely long straight wires A and B, e | vedclass

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(B) The magnetic field due to wire $A$ (along $-\hat{i}$ axis) at point $P(d, d)$ is given by the right-hand rule. The distance of point $P$ from the

Vedclass · Accepted Answer

$\frac{\mu_0 I q v}{\pi d} \hat{j}$

Vedclass · Answer

(B) The magnetic field due to wire $A$ (along $-\hat{i}$ axis) at point $P(d, d)$ is given by the right-hand rule. The distance of point $P$ from the $x$-axis is $d$. The magnetic field is $B_A = \frac{\mu_0 I}{2 \pi d} (-\hat{k})$.The magnetic field due to wire $B$ (along $\hat{j}$ axis) at point $P(d, d)$ is given by the right-hand rule. The distance of point $P$ from the $y$-axis is $d$. The magnetic field is $B_B = \frac{\mu_0 I}{2 \pi d} (-\hat{k})$.The resultant magnetic field at point $P$ is $B = B_A + B_B = \frac{\mu_0 I}{2 \pi d} (-\hat{k}) + \frac{\mu_0 I}{2 \pi d} (-\hat{k}) = \frac{\mu_0 I}{\pi d} (-\hat{k})$.The force on the charged particle is given by $F = q(v 	imes B)$.Given $v = v\hat{i}$ and $B = -\frac{\mu_0 I}{\pi d} \hat{k}$.$F = q(v\hat{i} 	imes (-\frac{\mu_0 I}{\pi d} \hat{k})) = -\frac{\mu_0 I q v}{\pi d} (\hat{i} 	imes \hat{k}) = -\frac{\mu_0 I q v}{\pi d} (-\hat{j}) = \frac{\mu_0 I q v}{\pi d} \hat{j}$.

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