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

Step 1: Understanding the Concept:
For a straight cylindrical wire carrying current, the magnetic field is zero at the center, increases linearly inside, and decreases as \( 1/r \) outside. The maximum field is always at the surface.
: Key Formula or Approach:
\[ B_{max} = \frac{\mu_0 I}{2\pi R} \]
where \( R \) is the radius of the wire.
Step 2: Detailed Explanation:
Given:
Diameter \( d = 1.6 \text{ mm} \implies R = 0.8 \text{ mm} = 0.8 \times 10^{-3} \text{ m} \).
\( B_{max} = 5 \times 10^{-4} \text{ T} \).
Substitute values:
\[ 5 \times 10^{-4} = \frac{4\pi \times 10^{-7} \times I}{2\pi \times (0.8 \times 10^{-3})} \]
Simplify the fraction:
\[ 5 \times 10^{-4} = \frac{2 \times 10^{-7} \times I}{0.8 \times 10^{-3}} \]
\[ I = \frac{5 \times 10^{-4} \times 0.8 \times 10^{-3}}{2 \times 10^{-7}} \]
\[ I = \frac{4 \times 10^{-7}}{2 \times 10^{-7}} = 2 \text{ A} \]
Step 3: Final Answer:
The value of current I is 2 A.