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Question

(C) The force per unit length between two parallel conductors carrying currents $I_1$ and $I_2$ separated by distance $d$ is given by $F = \frac{\mu_0

Vedclass · Accepted Answer

$-\frac{2F}{3}$

Vedclass · Answer

(C) The force per unit length between two parallel conductors carrying currents $I_1$ and $I_2$ separated by distance $d$ is given by $F = \frac{\mu_0 I_1 I_2}{2 \pi d}$.Since the currents are in the same direction,the force is attractive. Let the initial force be $F = k \frac{I_1 I_2}{d}$,where $k = \frac{\mu_0}{2 \pi}$.In the new scenario,the current in one conductor becomes $2I_1$ (or $2I_2$) and the direction is reversed,making the force repulsive (negative sign).The new distance is $d' = 3d$.The new force $F'$ is given by $F' = -k \frac{(2I_1) I_2}{3d} = -\frac{2}{3} \left( k \frac{I_1 I_2}{d} \right) = -\frac{2}{3} F$.

Point	Slope of the normal
$A. (7, 2)$	$1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$	$2. -8$
$C. (1, -1)$	$3. 4$
$D. (3, \sqrt{2})$	$4. 0$
	$5. -2\sqrt{2}$

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