Question

A solenoid of $2\text{m}$ long has $200$ turns per meter and current $2\text{A}$ flows through it as shown below. What would be the magnetic field at the point $\text{P}$ on the axis of the solenoid $5\text{m}$ away from the centre ?

• A. $1.40\times {10}^{-10}\text{T}$
• B. $1.04\times {10}^{-10}\text{T}$
• C. $0.14\times {10}^{-10}\text{T}$
• D. $0.70\times {10}^{-10}\text{T}$

Answer 1

The correct option is A $1.40\times {10}^{-10}\text{T}$

Given:
$2l=2\text{m};n=200;I=2\text{A}$
$x=5\text{m};A=1{\text{cm}}^2={10}^{-4}{\text{m}}^2$

As we know, field produced by a solenoid is similar to that of field produced by a bar magnet. Thus, the solenoid acts as a bar magnet having magnetic moment $M=NIA$.

Here,
$M=\text{magnetic moment}N=\text{total no of turns}A=\text{Area of the loops}$

$M=(200\times 2)\times 2\times {10}^{-4}$

$M=0.08{\text{Am}}^2$

Field along the axis of a solenoid is,

$B=\frac{{\mu }_0}{4\pi }\times \frac{2Mx}{(x^2-l^2{)}^2}$

$B={10}^{-7}\times \frac{2\times 0.08\times 5}{(5^2-1^2{)}^2}=\frac{0.8\times {10}^{-7}}{576}$

$\therefore B=1.4\times {10}^{-10}\text{T}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(A\right)$ is the correct answer.