A long solenoid carrying a current produces magnetic field B along its axis. If the number of turns per cm…
A long solenoid carrying a current produces magnetic field B along its axis. If the number of turns per cm are tripled and the current is made $\left(\frac{1}{4}\right)^{\mathrm{th}}$ then the new value of magnetic field will be
$\frac{B}{3}$
$\frac{\mathrm{B}}{4}$
$\frac{3 B}{4}$
$\frac{2 \mathrm{~B}}{3}$
Solution
Magnetic field due to a solenoid is given by, $B=\mu_0 \mathrm{NIL}$
After reducing current and increasing number of turns,
$\mathrm{B}^{\prime}=\mu_0(3 \mathrm{~N})\left(\frac{\mathrm{I}}{4}\right) \mathrm{L}=\frac{3}{4}\left(\mu_0 \mathrm{NIL}\right)=\frac{3}{4} \mathrm{~B}$