Question

A circular coil carrying a current I has radius R and number of turns N. If all the three, i.e. the current I, radius R and number of turns N are doubled, then, magnetic field at its centre becomes:

• A. One fourth
• B. Half
• C. Four times
• D. Double

Answer 1

Answer:

D. Double

The magnetic field (B) at the center of a circular coil is given by:

B = (μ₀ N I) / (2 * R)

where μ₀ is the magnetic constant.

If I, R, and N are all doubled:

B' = (μ₀ (2N) (2I)) / (2 * (2R))

B' = (μ₀ 2N 2I) / (4R)

B' = (μ₀ N I) / R

B' = 2 (μ₀ N * I) / (2R)

B' = 2B

So, the magnetic field at the center becomes double.