Question

A straight wire of length 20 cm carrying a current of \(\frac{3}{\pi ^2} \, \text{A} \) is bent in the form of a circle. The magnetic field at the centre of the circle is.

• A. \( 8 \times 10^{-6} \, \text{T} \)
• B. \( 3 \times 10^{-6} \, \text{T} \)
• C. \( 12 \times 10^{-6} \, \text{T} \)
• D. \( 6 \times 10^{-6} \, \text{T} \) \bigskip

Hint:

Use the formula for the magnetic field at the center of a current-carrying loop to find the field at any point.

Answer 1

The Correct Option is D

Step 1: Determining the Radius of the Circle The length of the wire is given as: \[ L = 20 \text{ cm} = 0.2 \text{ m} \] Since the wire is bent into a complete circle, the circumference of the circle is: \[ 2\pi R = L \] \[ R = \frac{L}{2\pi} = \frac{0.2}{2\pi} = \frac{0.1}{\pi} \text{ m} \] 

 Step 2: Magnetic Field at the Centre of a Circular Loop The magnetic field at the center of a current-carrying circular loop is given by: \[ B = \frac{\mu_0 I}{2R} \] where: - \( \mu_0 = 4\pi \times 10^{-7} \text{ T m/A} \) (permeability of free space) - \( I = \frac{3}{\pi^2} \) A (current in the wire) - \( R = \frac{0.1}{\pi} \) m (radius of the circle) 

Step 3: Substituting Values \[ B = \frac{(4\pi \times 10^{-7}) \times \left(\frac{3}{\pi^2}\right)}{2 \times \frac{0.1}{\pi}} \] Simplifying: \[ B = \frac{4\pi \times 3 \times 10^{-7}}{2\pi \times \frac{0.1}{\pi}} \] \[ B = \frac{12\pi \times 10^{-7}}{0.2} \] \[ B = \frac{12 \times 10^{-7} \times \pi}{0.2} \] \[ B = 6 \times 10^{-6} \text{ T} \] Thus, the correct answer is: \[ \mathbf{6 \times 10^{-6} \, \text{T}} \]