Question

A copper wire of diameter 1.6 mm carries a current I. The maximum magnetic field due to this wire is \(5 \times 10^{-4} \, T\). The value of I is :

• A. 0.2 A
• B. 0.5 A
• C. 2 A
• D. 4 A

Hint:

Maximum magnetic field for a wire occurs at its surface.

Answer 1

The Correct Option is C

Concept: Magnetic field near surface of wire: \[ B = \frac{\mu_0 I}{2\pi r} \]

Step 1: Given values.
\[ d = 1.6 \, \text{mm} \Rightarrow r = 0.8 \times 10^{-3} \, m \] \[ \mu_0 = 4\pi \times 10^{-7} \]

Step 2: Substitute.
\[ 5 \times 10^{-4} = \frac{4\pi \times 10^{-7} \cdot I}{2\pi \cdot 0.8 \times 10^{-3}} \]

Step 3: Solve.
\[ 5 \times 10^{-4} = \frac{2 \times 10^{-7} I}{0.8 \times 10^{-3}} = \frac{2I}{0.8} \times 10^{-4} \] \[ 5 = \frac{2I}{0.8} \Rightarrow I = 2 \, A \]