Question

A long solenoid carrying a current produces a magnetic field $B$ along its axis. If the current is doubled and the number of turns per $cm$ is halved, then the new value of the magnetic field is

• A. $B$
• B. $2B$
• C. $4B$
• D. $B / 2$

Answer 1

Answer: (A)

Magnetic field on the axis of a solenoid
$B=\mu _{0}ni$
$B′=\left(\mu \right)_{0}\left(\frac{n}{2}\right)\left(2 i\right)=B$