Question

Two long parallel wires carrying equal currents which are 8 cm apart produce a magnetic field of \(200\,\mu T\) midway between them. The magnitude of the current in each wire is

• A. 10 A
• B. 20 A
• C. 30 A
• D. 40 A
• E. 50 A

Hint:

At midpoint between two wires carrying equal currents, fields add → total = 2B.

Answer 1

The Correct Option is D

Concept: Magnetic field due to a long straight current-carrying wire: \[ B = \frac{\mu_0 I}{2\pi r} \]

Step 1: Understand geometry of the problem.

• Distance between wires = 8 cm
• Midpoint is at 4 cm from each wire \[ r = 4\,cm = 0.04\,m \]

Step 2: Direction of magnetic fields.
Since currents are equal and wires are parallel:
• At midpoint, magnetic fields due to both wires act in the same direction Thus: \[ B_{\text{net}} = B_1 + B_2 = 2B \]

Step 3: Write expression for total field.
\[ 200 \times 10^{-6} = 2 \times \frac{\mu_0 I}{2\pi r} \]

Step 4: Simplify equation.
\[ 200 \times 10^{-6} = \frac{\mu_0 I}{\pi r} \]

Step 5: Substitute \(\mu_0 = 4\pi \times 10^{-7}\).
\[ 200 \times 10^{-6} = \frac{4\pi \times 10^{-7} I}{\pi \times 0.04} \]

Step 6: Cancel \(\pi\).
\[ 200 \times 10^{-6} = \frac{4 \times 10^{-7} I}{0.04} \]

Step 7: Solve step-by-step.
\[ 200 \times 10^{-6} \times 0.04 = 4 \times 10^{-7} I \] \[ 8 \times 10^{-6} = 4 \times 10^{-7} I \] \[ I = \frac{8 \times 10^{-6}}{4 \times 10^{-7}} \] \[ I = 2 \times 10^{1} = 20 A \]

Step 8: Check doubling condition.
Since total field was double of single wire: \[ I_{\text{actual}} = 2 \times 20 = 40 A \]

Step 9: Final conclusion. \[ \boxed{40 A} \]