Question

The length of the solenoid is $0.1\text{m}$ and its diameter is very small. A wire is wound over it in two layers. The number of turns in inner layer is $50$ and that in outer layer is $40$. The strength of current flowing in two layers in opposite direction is $3\text{A}$. Then the magnetic induction at the middle of solenoid is:

• A. $12\pi \times {10}^{-5}\text{T}$
• B. $3\pi \times {10}^{-5}\text{T}$
• C. $7\pi \times {10}^{-5}\text{T}$
• D. $9\pi \times {10}^{-5}\text{T}$

Answer 1

The correct option is A $12\pi \times {10}^{-5}\text{T}$

The diameter of solenoid is very small w.r.t. its length $l(l>>r)$, hence it is an ideal solenoid.

The magnetic field due to both layers at the centre of solenoid will be in opposite direction, because the direction of current in both layers is in opposite direction.

$\therefore B_{net}=B_1-B_2$

$B_{net}={\mu }_0{n}_1{i}_1-{\mu }_0{n}_2{i}_2$

$B_{net}={\mu }_0\left(\frac{N_1}{l}\right)i-{\mu }_0\left(\frac{N_2}{l}\right)i[\because i_1=i_2=i]$

$B_{net}=\frac{{\mu }_{0}i}{l}(N_1-N_2)$

Here,

$i=3\text{A};l=0.1\text{m};N_1=50;N_2=40$

$B_{net}=\frac{4\pi \times {10}^{-7}\times 3\times (50-40)}{0.1}$

$\therefore B_{net}=12\pi \times {10}^{-5}\text{T}$

Hence, option $\left(a\right)$ is correct.