Question

If we double the radius of current carrying coil keeping the current unchanged, what happens to the magnetic field at centre?
1. become four times
2. doubled
3. remains unchanged
4. halved

Answer 1

Correct Answer - Option 4 : halved

The correct answer is option 4) i.e. halved.

CONCEPT:

• Magnetic field due to a current-carrying conductor: Biot-Savart Law

​Magnetic field B at a radial distance r, due to a wire carrying current coil is given by:

\(B = \frac{μ_0I}{2π r}\)B=μ0I2πrB=μ0I2πr

Where μ0 is the permeability of free space (4π × 10-7 Tm/A), and I is the current intensity.

EXPLANATION:

Let the magnetic field due to a current-carrying coil of radius R be B. 

⇒\(B = \frac{μ_0I}{2πR}\)

When the radius of the coil is doubled, R' = 2R

Therefore, the magnetic field at the center B' = \(\frac{μ_0I}{2πR'} = \frac{μ_0I}{2π(2R)} =\frac{1}{2}B\)

Hence, the magnetic field is halved as the radius of the coil is doubled.