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

The problem involves two long parallel wires carrying current. We need to calculate the change in magnetic force when the configuration is altered. Let's solve this step-by-step:

1. For two long parallel wires carrying currents \(I_1\) and \(I_2\), separated by a distance \(d\), the magnetic force per unit length \(f\) that each wire experiences is given by the formula: \(f = \frac{\mu_0}{2\pi} \cdot \frac{I_1 \cdot I_2}{d}\), where \(\mu_0\) is the permeability of free space.
2. Initial conditions:
   • Currents: \(I_1 = 10\, A\), \(I_2 = 10\, A\)
   • Distance: \(d = 5\, cm = 0.05\, m\)
   • Length of wire considered: \(L = 10\, cm = 0.1\, m\)
3. The initial force per unit length \(f_1\) is then: \(f_1 = \frac{\mu_0}{2\pi} \cdot \frac{10 \times 10}{0.05}\)
4. The total force on the \(10\, cm\) length of wire \(P\), denoted as \(F_1\), is: \(F_1 = f_1 \times 0.1\)
5. In the new scenario:
   • Distance is halved: \(d' = \frac{0.05}{2} = 0.025\, m\)
   • Currents are doubled: \(I_1' = I_2' = 20\, A\)
6. Calculating the new force per unit length \(f_2\): \(f_2 = \frac{\mu_0}{2\pi} \cdot \frac{20 \times 20}{0.025}\)
7. The new total force on the \(10\, cm\) length of wire \(P\), denoted as \(F_2\), is: \(F_2 = f_2 \times 0.1\)
8. Solving these equations, we can calculate the factor by which the force increases: \[ \frac{F_2}{F_1} = \frac{f_2}{f_1} = \frac{20^2/0.025}{10^2/0.05} = 8 \]
9. Thus, the force increases by a factor of 8. Therefore, \(F_2 = 8F_1\).

The correct answer is \(8F_1\).