Question

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

• A. $\frac{\mathrm{B}}{2}$
• B. $\mathrm{B}$
• C. $2\mathrm{B}$
• D. $4\mathrm{B}$

Answer 1

The correct option is B

$\mathrm{B}$

Step 1: Given data

A long solenoid carrying a current produces a magnetic field $\mathrm{B}$ along its axis.

The current is doubled, that is $2\mathrm{I}$.

The number of turns per cm is halved, that is $\frac{\mathrm{N}}{2}$.

Step 2: Formula used

The formula for the magnetic field of a solenoid is,

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{IN}}{\mathrm{L}}.....\left(1\right)$

Where,

${\mathrm{\mu }}_0$ is the permeability in free space

$\mathrm{N}$ is the number of turns in the solenoid

$\mathrm{I}$ is current in the coil

$\mathrm{L}$ is the length of the coil.

Step 3: New value of the magnetic field:

Here,

The current is doubled, that is $2\mathrm{I}$

The number of turns per cm is halved, that is $\frac{\mathrm{N}}{2}$

The new magnetic field($\mathrm{B}\text{'}$) of a solenoid is,

$$\begin{gathered} \mathrm{B}\text{'}=\frac{{\mathrm{\mu }}_{0}2\mathrm{I}{\displaystyle \frac{\mathrm{N}}{2}}}{\mathrm{L}} \\ \mathrm{B}\text{'}=\frac{{\mathrm{\mu }}_0\mathrm{I}{\displaystyle \mathrm{N}}}{\mathrm{L}}.....\left(2\right) \end{gathered}$$

Hence, equations (1) and (2) are the same

That is $\mathrm{B}=\mathrm{B}\text{'}$

Therefore, the current is doubled and the number of turns per cm is halved. the new value of the magnetic field is $\mathrm{B}$.

Hence, option B is correct.