Question

A rectangular coil of length 10 cm and breadth 9 cm carries a current of 10 A. A long straight conductor carrying a current of 20 A is placed 1 cm from the coil parallel to its length and in the same plane of the coil. What will be the net force acting on the straight conductor?

• A. \( 1.8 \times 10^{-4} \, N \)
• B. \( 3.6 \times 10^{-4} \, N \)
• C. \( 1.8 \times 10^{-5} \, N \)
• D. \( 3.6 \times 10^{-5} \, N \)

Hint:

Opposite currents on opposite sides create difference in forces → net force = difference of two.

Answer 1

The Correct Option is B

Step 1: Force between parallel currents.
\[ F = \frac{\mu_0 I_1 I_2 l}{2\pi r} \]

Step 2: Forces due to two sides.
Near side: distance \( r_1 = 0.01 \, m \)
Far side: \( r_2 = 0.01 + 0.09 = 0.10 \, m \)

Step 3: Net force expression.
\[ F = \frac{\mu_0 I_1 I_2 l}{2\pi} \left(\frac{1}{r_1} - \frac{1}{r_2}\right) \]

Step 4: Substitute values.
\[ \mu_0 = 4\pi \times 10^{-7}, \; I_1=10, \; I_2=20, \; l=0.1 \]

Step 5: Simplify.
\[ F = \frac{4\pi \times 10^{-7} \cdot 200 \cdot 0.1}{2\pi} \left(100 - 10\right) \]

Step 6: Calculation.
\[ F = 2 \times 10^{-7} \cdot 20 \cdot 90 = 3.6 \times 10^{-4} N \]

Step 7: Final Answer.
\[ \boxed{3.6 \times 10^{-4} \, N} \]