Question

A long solenoid has 200 turns per cm and carries a current $I$. The magnetic field at its centre is $6.28 \times {10^{ - 2}}\,{\text{Wb/}}{{\text{m}}^2}$. Another long solenoid has 100 turns per cm and it carries a current $I/3$. The value of the magnetic field at its centre is
A. $1.05 \times {10^{ - 2}}\,{\text{Wb/}}{{\text{m}}^2}$
B. $1.05 \times {10^{ - 5}}\,{\text{Wb/}}{{\text{m}}^2}$
C. $1.05 \times {10^{ - 3}}\,{\text{Wb/}}{{\text{m}}^2}$
D. $1.05 \times {10^{ - 4}}\,{\text{Wb/}}{{\text{m}}^2}$

Answer 1

Use the formula for the magnetic field inside the solenoid. This formula gives the relation between the magnetic field at the centre of the solenoid, number of turns per unit length of the solenoid and the current flowing through the coil of solenoid. Rewrite this equation for two solenoids given and solve it for the magnetic field at the centre of another solenoid.

Formula used:
The magnetic field in a solenoid is given by
$B = {\mu _0}nI$ …… (1)
Here, $B$ is the magnetic field in the solenoid, ${\mu _0}$ is the permeability of the free space, $n$ is the number of turns per unit length of the solenoid and $I$ is the current flowing in the coil of the solenoid.

Complete step by step answer:
We have given that the first long solenoid has 200 turns per cm and the current flowing through the coil of the solenoid is $I$. The magnetic field at the centre of this solenoid is $6.28 \times {10^{ - 2}}\,{\text{Wb/}}{{\text{m}}^2}$.
${n_1} = 200\,{\text{turns/cm}}$
${I_1} = I$
${B_1} = 6.28 \times {10^{ - 2}}\,{\text{Wb/}}{{\text{m}}^2}$
The current flowing through another solenoid is $I/3$ and the number of turns are 100 turns per cm.
${I_2} = \dfrac{I}{3}$
${n_2} = 100\,{\text{turns/cm}}$
Rewrite equation (1) for the magnetic field inside the first solenoid.
${B_1} = {\mu _0}{n_1}{I_1}$ …… (2)
Rewrite equation (1) for the magnetic field inside the another solenoid.
${B_2} = {\mu _0}{n_2}{I_2}$ …… (3)
Divide equation (3) by equation (2).
$\dfrac{{{B_2}}}{{{B_1}}} = \dfrac{{{\mu _0}{n_2}{I_2}}}{{{\mu _0}{n_1}{I_1}}}$
$\Rightarrow \dfrac{{{B_2}}}{{{B_1}}} = \dfrac{{{n_2}{I_2}}}{{{n_1}{I_1}}}$
Rearrange the above equation for ${B_2}$.
$\Rightarrow {B_2} = \dfrac{{{n_2}{I_2}{B_1}}}{{{n_1}{I_1}}}$
Substitute $100\,{\text{turns/cm}}$ for ${n_2}$, $\dfrac{I}{3}$ for ${I_2}$, $6.28 \times {10^{ - 2}}\,{\text{Wb/}}{{\text{m}}^2}$ for ${B_1}$, $200\,{\text{turns/cm}}$ for ${n_1}$ and $I$ for ${I_1}$ in the above equation.
$\Rightarrow {B_2} = \dfrac{{\left( {100\,{\text{turns/cm}}} \right)\left( {\dfrac{I}{3}} \right)\left( {6.28 \times {{10}^{ - 2}}\,{\text{Wb/}}{{\text{m}}^2}} \right)}}{{\left( {200\,{\text{turns/cm}}} \right)I}}$
$\Rightarrow {B_2} = 1.046 \times {10^{ - 2}}\,{\text{Wb/}}{{\text{m}}^2}$
$\Rightarrow {B_2} = 1.05 \times {10^{ - 2}}\,{\text{Wb/}}{{\text{m}}^2}$
Therefore, the magnetic field at the centre of another solenoid is $1.05 \times {10^{ - 2}}\,{\text{Wb/}}{{\text{m}}^2}$.

So, the correct answer is “Option A”.

Note:
The students may convert the unit of the number of turns per unit length of the solenoid. But there is no need for this unit conversion as units of the term number of turns per unit length will get cancelled in the calculation. So, it will be a waste of time to convert the units in the SI system of units.