A rectangular loop ABCD made of uniform wire is a part of an electric circuit shown in the given figure. The length AD = BC= 5 mm. The sides AB and DC are long compared to the other two sides. What will be the magnetic force per unit length acting on the wire DC due to the wire AB if the ammeter reading is 8 A?

$6.4 × 10^{-4} Nm^{-1}$

The correct answer is Option (2) → $6.4 × 10^{-4} Nm^{-1}$

Given:

Total current from ammeter $I_{\text{total}} = 8\ \text{A}$

Separation between parallel wires $r = AD = 5\ \text{mm} = 5\times10^{-3}\ \text{m}$

$\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$

Note: The top wire AB and bottom wire DC form two parallel branches between the same nodes. The total current divides equally into these two identical branches, so

$I_{AB}=I_{DC}=\frac{I_{\text{total}}}{2}=\frac{8}{2}=4\ \text{A}$

Force per unit length between two long parallel currents:

$\displaystyle \frac{F}{\ell}=\frac{\mu_0 I_1 I_2}{2\pi r}$

Substitute $I_1=I_2=4\ \text{A}$:

$\displaystyle \frac{F}{\ell}=\frac{4\pi\times10^{-7}\times 4 \times 4}{2\pi \times 5\times10^{-3}} =\frac{4\pi\times10^{-7}\times 16}{2\pi\times5\times10^{-3}}$

$\displaystyle \frac{F}{\ell}=\frac{64\times10^{-7}}{10\times10^{-3}}=6.4\times10^{-4}\ \text{N·m}^{-1}$

Therefore, $ \displaystyle \frac{F}{\ell}=6.4\times10^{-4}\ \text{N·m}^{-1}$ (repulsive).