Question

A winding wire which is used to frame a long solenoid can withstand a maximum current of$10\text{A}$. If the length of solenoid is $80\text{cm}$ and its cross- sectional radius is $3\text{cm}$, then find the required length of the winding wire if the magnetic field inside the solenoid near its axis is $0.2\text{T}$.

• A. $1.2\times {10}^2\text{m}$
• B. $4.8\times {10}^2\text{m}$
• C. $2.4\times {10}^3\text{m}$
• D. $6\times {10}^3\text{m}$

Answer 1

The correct option is C $2.4\times {10}^3\text{m}$

The magnetic field inside the long solenoid is given by

$B={\mu }_{0}ni={\mu }_0(\frac{N}{l})i$

Where, $n=\frac{N}{l}=$ number of turns per unit length

Given data:
$i=10\text{A};l=0.8\text{A};B=0.2\text{T}$

$0.2=\frac{4\pi \times {10}^{-7}\times N\times 10}{0.80}$

$\therefore N=\frac{4\times {10}^4}{\pi }$

Since the $N$ turns are made from the winding wire hence the length $l$ of wire is

$l=2\pi r\times N$

$l=2\pi \times (3\times {10}^{-2})\times \frac{4\times {10}^4}{\pi }$

$\therefore l=2400=2.4\times {10}^3\text{m}$

Therefore, option $\left(c\right)$ is correct.

Why this question? Note: The total length of wire required for winding is equal to the perimeter of the solenoid which is $(N\times 2\pi r)$