Question

There are 50 turns of a wire in every $1cm$ length of a long solenoid. If a $4A$ current is flowing in the solenoid, the approximate value of magnetic field along its axis at an internal point and at one end will be respectively

• A. $12.6\times {10}^{-3}\frac{Weber}{m^2},6.3\times {10}^{-3}\frac{Weber}{m^2}$
• B. $12.6\times {10}^{-3}\frac{Weber}{m^2},25.1\times {10}^{-3}\frac{Weber}{m^2}$
• C. $25.1\times {10}^{-3}\frac{Weber}{m^2},12.6\times {10}^{-3}\frac{Weber}{m^2}$
• D. $25.1\times {10}^{-5}\frac{Weber}{m^2},6.3\times {10}^{-5}\frac{Weber}{m^2}$

Answer 1

The correct option is C $25.1\times {10}^{-3}\frac{Weber}{m^2},12.6\times {10}^{-3}\frac{Weber}{m^2}$

No of turns per metre $=\frac{50}{0.01}=5000$
Magnetic field inside the solenoid is given by
$B_{in}={\mu }_{0}ni$

and magnetic field at the ends of a solenoid is given by
$B_{end}=\frac{B_{in}}{2}=\frac{{\mu }_{0}ni}{2}$

Substituting the values of ${\mu }_0=4\pi \times {10}^{-7},i=4A$ and $n=5000$, we get
$B_{in}=25.1\times {10}^{-3}\frac{Weber}{m^2}$ and $B_{end}=12.6\times {10}^{-3}\frac{Weber}{m^2}$