Question

A long solenoid carrying a current produces a magnetic field $B$ 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. $\dfrac{B}{2}$
B. $B$
C. $2B$
D. $4\,B$

Answer 1

Hint: The magnetic field of a solenoid depends directly on the current flowing through the solenoid and the number of turns. Using this find the final magnetic field when the number of turns is halved and current is doubled. On comparing it with the initial magnetic field we can find the final answer.

Complete step by step answer:
It is given that a solenoid carrying a current produces a magnetic field $B$ along its axis.
We need to find the magnetic field when the current in the solenoid is doubled and the number of turns per cm is halved.
We know that the magnetic field in a solenoid depends directly on the current and number of turns.
The equation for finding magnetic field induction for a solenoid is given as
$B = {\mu _0}ni$
Where, ${\mu _0}$ is the permeability in free space, $i$ is the current flowing through it and $n$ is the number of turns of the coil.
Let the initial current be ${i_1}$
and the number of turns be ${n_1}$
Then the magnetic field induction will be
$ \Rightarrow B = {\mu _0}{n_1}{i_1}$
Now let us calculate the final magnetic field.
Let the new current be ${i_2}$ and the number of turns be ${n_2}$
It is given that new current is twice that of initial current.
$ \Rightarrow {i_2} = 2{i_1}$
Number of turns is half the initial value.
$ \Rightarrow {n_2} = \dfrac{{{n_1}}}{2}$
So, the new magnetic field is given as
$ \Rightarrow B' = {\mu _0}{n_2}{i_2}$
$ \Rightarrow B' = {\mu _0}\dfrac{{{n_1}}}{2}2{i_1}$
$ \Rightarrow B' = {\mu _0}{n_1}{i_1}$
This value is the same as that of the initial magnetic field.
$ \Rightarrow B' = B$
Which means there is no change in magnetic field if you double the current and reduce the number of turns to half.

Hence the correct answer is option B.

Note:
In the case of a current carrying solenoid the maximum magnetic field will be along the axis passing through its centre. It is calculated as ${\rm B} = {\mu _0}ni$, where $n$ is the number of turns and $i$ is the current. The value of the magnetic field through this axis will be constant. Whereas outside a solenoid the magnetic field is zero.