Q.
Force per unit length between long parallel wires in the circuit is $3.6 \times 10^{-3} \, N \, m^{-1}$. Resistance of the circuit is
Solution:
Force per unit length between the parallel wires is given by,
$F = \frac{\mu_0}{2 \pi} \frac{I_1 I_2}{d}$
Here, $F = 3.6 \times 10^{-3} N \, m^{-1} , I_1 = I_2 = I $
$d= 8 mm = 8 \times 10^{-3} m $
So, $3.6 \times 10^{-3} = 2 \times 10^{-7} \times \frac{I^2}{8 \times 10^{-3}}$
or, $I^2 = 40 \times 3.6 $
or, $I = 12 \, A $
Resistance of the circuit,
$R = \frac{V}{I} = \frac{18}{12} = 1.5 \, Omega$
