Question

A long solenoid is fabricated by closely winding a wire of radius $0.5\text{mm}$ over a cylindrical non-magnetic frame such that the successive turns nearly touch each other. What would be the magnetic field $B$ at the center of the solenoid if it carries a current of $5\text{A}$ ?

• A. $3.5\pi \times {10}^{-3}\text{T}$
• B. $2\pi \times {10}^{-3}\text{T}$
• C. $5\pi \times {10}^{-3}\text{T}$
• D. $\pi \times {10}^{-3}\text{T}$

Answer 1

The correct option is B $2\pi \times {10}^{-3}\text{T}$

Let the successive turns of solenoid touch each other as shown in figure.

Width of each turn = diameter $d$ of wire

$\therefore$ width of each turn $=d=2r=1\text{mm}$

Width of each turn $=1\times {10}^{-3}\text{m}$

Thus, the number of turns per unit length of solenoid will be

$n=\frac{1\text{m}}{d}=\frac{1\text{m}}{1\times {10}^{-3}\text{m}}=1000$

current in solenoid is $i=5\text{A}$

Using formula, $B={\mu }_{0}ni$

$B=4\pi \times {10}^{-7}\times 1000\times 5=20\pi \times {10}^{-4}$

$\therefore B=2\pi \times {10}^{-3}\text{T}$

Hence, option $\left(b\right)$ is the correct answer.