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

Answer 1

Answer: (A)

Double

Magnetic field at the center of coil is given by $B=\frac{\mu NI}{2R}$
New current $I^{\prime }=2I$
New radius $R^{\prime }=2R$
New number of turns $N^{\prime }=2N$
Thus, magnetic field $B^{\prime }=\frac{{\mu }_o{N}^{\prime }{I}^{\prime }}{2R^{\prime }}=\frac{{\mu }_o\left(2N\right)\left(2I\right)}{2\left(2R\right)}$
$⟹B^{\prime }=2\times \frac{{\mu }_{o}NI}{2R}=2B$
Thus magnetic field gets doubled.