Two parallel conducting wires ‘A’ and ‘B’ carrying currents are separated by \( 0.2 \) m. The force on \( 8 \) m segment of ‘A’ due to ‘B’ is \( 10 \) N. If length of ‘B’ is doubled and current through it is tripled, force on ‘B’ per unit length will become
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 3.75 \) N
The force per unit length between two long parallel currents is the same on each wire (Newton’s third law):
\( \displaystyle \frac{F}{l} = \frac{\mu_0I_AI_B}{2\pi d} \)
From the given data, the force per unit length at present is
\( \displaystyle \frac{F}{l} = \frac{10\,\mathrm{N}}{8\,\mathrm{m}} = 1.25\,\mathrm{N\,m^{-1}} \)
F/l is proportional to \( I_B \), so tripling \( I_B \) makes it three times. The length of B does not enter the formula for long wires, so doubling the length changes nothing.
\( \frac{F}{l} = 3\times1.25 = 3.75\,\mathrm{N\,m^{-1}} \)
Note: It seems that there is a mistake in the original NTA paper, which prints the options in N when the force per unit length should be in N m−1. The question asks for the force per unit length, so the answer 3.75 is read as 3.75 N m−1, and the question can still be solved.