Question

Two long straight wires $P$ and $Q$ carrying equal current $10\, A$ each were kept parallel to each other at $5 \, cm$ distance Magnitude of magnetic force experienced by $10 \, cm$ length of wire $P$ is $F_1$ If distance between wires is halved and currents on them are doubled, force $F_2$ on $10 \, cm$ length of wire $P$ will be:

• A. $10 F _1$
• B. $8 F _1$
• C. $\frac{F_1}{8}$
• D. $\frac{F_1}{10}$

Answer 1

The Correct Option is B

Force per unit length between two parallel straight wires:

The force per unit length between two parallel wires carrying currents \( i_1 \) and \( i_2 \) is given by the formula:

\[ F = \frac{\mu_0 i_1 i_2}{2\pi d} \]

Given that the currents in two cases are \( i_1 = 10 \) A and \( i_2 = 20 \) A, and the separation distance is \( d = 5 \) cm in the first case and \( d = \frac{5}{2} \) cm in the second case, the force ratio can be calculated as:

\[ \frac{F_1}{F_2} = \frac{\frac{\mu_0 (10)^2}{2\pi (5cm)}}{\frac{\mu_0 (20)^2}{2\pi \left(\frac{5cm}{2}\right)}} \]

Simplifying the expression:

\[ \frac{F_1}{F_2} = \frac{100}{5} \times \frac{1}{\left(\frac{400}{\frac{5}{2}}\right)} \]

\[ = \frac{100}{5} \times \frac{2}{400} \]

\[ = \frac{100 \times 2}{5 \times 400} \]

\[ = \frac{200}{2000} = \frac{1}{8} \]

Therefore, we get:

\[ F_2 = 8F_1 \]