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?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 6.4 \times 10^{-4}\,\mathrm{N\,m^{-1}} \)
The current enters at the middle of AD and leaves at the middle of BC. The wire is uniform, so the two paths (through AB and through DC) have equal resistance and the 8 A divides equally: 4 A flows in AB and 4 A in DC, in the same direction.
AB and DC are long compared with their separation, so they act as long parallel wires:
\( \displaystyle \frac{F}{l} = \frac{\mu_0\,I_1 I_2}{2\pi d} \)
\( I_1 = I_2 = 4\,\mathrm{A} \)
\( d = 5\,\mathrm{mm} = 5\times10^{-3}\,\mathrm{m} \)
\( \mu_0/2\pi = 2\times10^{-7}\,\mathrm{T\,m\,A^{-1}} \)
\( \displaystyle \frac{F}{l} = \frac{2\times10^{-7} \times 4 \times 4}{5\times10^{-3}} = 6.4\times10^{-4}\,\mathrm{N\,m^{-1}} \)
The currents are parallel, so the force is attractive.