Question

An element of \( 0.05 \) m is placed at the origin, carrying a large current of \( 10 A \). The magnetic field at a perpendicular distance of \( 1 \) m is: 
 

• A. \( 4.5 \times 10^{-8} \) T
• B. \( 5.5 \times 10^{-8} \) T
• C. \( 5.0 \times 10^{-8} \) T
• D. \( 7.5 \times 10^{-8} \) T

Hint:

For magnetic fields due to current elements:
- The Biot-Savart law applies for small segments.
- The field direction is given by the right-hand rule.
- \( dB \) is maximum when the element is perpendicular to the position vector.

Answer 1

The Correct Option is C

Step 1: Using the Biot-Savart law for a small current element:
\[ dB = \frac{\mu_0}{4\pi} \frac{I dl \sin\theta}{r^2} \] where \( \mu_0 = 4\pi \times 10^{-7} \), \( I = 10 A \), \( dl = 0.05 m \), and \( r = 1 m \).
Step 2: Since the element is perpendicular to the position vector, \( \sin\theta = 1 \):
\[ dB = \frac{(4\pi \times 10^{-7}) (10) (0.05)}{4\pi (1)^2} \]
Step 3: Simplifying:
\[ dB = \frac{10^{-7} \times 10 \times 0.05}{1} = 5.0 \times 10^{-8} T \]