A long solenoid carrying a current produces a magnetic field B along its axis. If the number of turns per…
A long solenoid carrying a current produces a magnetic field B along its axis. If the number of turns per $\mathrm{cm}$ is doubled and the current is made, $\left(\frac{1}{3}\right)^{\text {rd }}$ then the new value of the magnetic field will be
$\frac{B}{3}$
$3 \mathrm{~B}$
$2 \mathrm{~B}$
$\frac{2 \mathrm{~B}}{3}$
Solution
Magnetic field inside a solenoid is given by
$\mathrm{B}=\mu_0 \mathrm{nI}$
If $\mathrm{n}$ is doubled and $\mathrm{I}$ is made $\left(\frac{1}{3}\right)^{\mathrm{rd}}$, the value of $\mathrm{B}$ will become $\frac{2}{3} \mathrm{~B}$