Question

What is the strength of the new magnetic field due to the solenoid, if the number of turns are reduced to $(1/4{)}^{\text{th}}$ of initial turns and current is tripled ?

• A. $2$ times of initial value of the magnetic field strength
• B. $0.75$ times of initial value of the magnetic field strength
• C. $3$ times of initial value of the magnetic field strength
• D. $4$ times of initial value of the magnetic field strength

Answer 1

The correct option is B $0.75$ times of initial value of the magnetic field strength

Given:
$n_2=\frac{1}{4}{n}_1;i_2=3i_1$

Magnetic field at the centre of the solenoid is given by

$B={\mu }_{0}ni$

$\therefore B_1={\mu }_0{n}_1{i}_1$ and $B_2={\mu }_0{n}_2{i}_2$

$\frac{B_2}{B_1}=\frac{{\mu }_0{n}_2{i}_2}{{\mu }_0{n}_1{i}_1}=(\frac{n_2}{n_1})(\frac{i_2}{i_1})$

$\frac{B_2}{B_1}=(\frac{\left(1/4\right)n_1}{n_1})(\frac{3i_1}{i_1})=\frac{3}{4}$

$\therefore B_2=0.75B_1$

Therefore, option $\left(b\right)$ is the correct answer.

Why this Question? Key point- The magnetic field due to an ideal solenoid is given by, $B={\mu }_{0}ni$ $\Rightarrow B\propto n,B\propto i$