Question

Keeping the magnetic field constant, if the number of turns are halved, what is the value of new current through the solenoid when a current of $2.5\text{A}$ was flowing initially?

• A. $1.25\text{A}$
• B. $7.5\text{A}$
• C. $0.5\text{A}$
• D. $5\text{A}$

Answer 1

The correct option is D $5\text{A}$

Given,

$B_1=B_2$

$n_2=\frac{1}{2}{n}_1$

$i_1=2.5\text{A},i_2=?$

For a solenoid,$B={\mu }_{0}ni$

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

$\because B_1=B_2$

${\mu }_0{n}_1{i}_1={\mu }_0{n}_2{i}_2$

So,

$i_2=\frac{n_1{i}_1}{n_2}=\frac{n_2\times 2.5}{\frac{1}{2}{n}_1}$

$\Rightarrow i_2=5\text{A}$

Hence, option $\left(d\right)$ is the correct answer.

Why this Question ? Note: For a given solenoid, the magnetic field is given by, $B={\mu }_{0}ni$ If $B$ remains same, $ni=$ constant $\Rightarrow n_1{i}_1=n_2{i}_2$