Question

The strength of earth's magnetic field at a point is 0.4×10^-5 T. If this field is to be annulled by the magnetic induction produced at the centre of a circular conducting loop of radius π cm, the current to be sent through the loop is

• A. 2A
• B. 0.15A
• C. 1.5A
• D. 0.2A
• E. 1A

Answer 1

The Correct Option is D

The magnetic field at the center of a circular loop carrying a current \( I \) is given by the formula: \[ B = \frac{\mu_0 I}{2r} \] where: 
\( B \) is the magnetic field at the center,
\( \mu_0 \) is the permeability of free space (\( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)),
\( I \) is the current,
\( r \) is the radius of the loop. We are given:
\( B = 0.4 \times 10^{-5} \, \text{T} \),
\( r = \pi \, \text{cm} = \frac{\pi}{100} \, \text{m} \).

Now, we solve for \( I \): \[ 0.4 \times 10^{-5} = \frac{4\pi \times 10^{-7} I}{2 \times \frac{\pi}{100}} \] Simplifying: \[ 0.4 \times 10^{-5} = \frac{2 \times 10^{-7} I}{\frac{1}{100}} \] \[ 0.4 \times 10^{-5} = 2 \times 10^{-5} I \] Solving for \( I \): \[ I = \frac{0.4 \times 10^{-5}}{2 \times 10^{-5}} = 0.2 \, \text{A} \] Thus, the current to be sent through the loop is \( 0.2 \, \text{A} \), which is option (D).

Answer 2

Given: 

• Magnetic field to be annulled: \( B = 0.4 \times 10^{-5} \, \text{T} \)
• Radius of the loop: \( r = \pi \, \text{cm} = \pi \times 10^{-2} \, \text{m} \)

Magnetic field at the center of a circular loop is given by: 

\( B = \frac{\mu_0 I}{2r} \) 
Rearranging to find current \( I \): 
\( I = \frac{2Br}{\mu_0} \) 

Substituting the values: 
\( \mu_0 = 4\pi \times 10^{-7} \, \text{T.m/A} \) \( B = 0.4 \times 10^{-5} \, \text{T} \) \( r = \pi \times 10^{-2} \, \text{m} \) 

\( I = \frac{2 \times 0.4 \times 10^{-5} \times \pi \times 10^{-2}}{4\pi \times 10^{-7}} \) 

Simplifying: 
\( I = \frac{0.8 \times 10^{-7} \pi}{4\pi \times 10^{-7}} = \frac{0.8}{4} = 0.2 \, \text{A} \) 

Final Answer: \( \boxed{0.2 \, \text{A}} \)