Question

A solenoid has $2\times {10}^4$ turns/meter and has diameter 10 cm. An electron beam having K.E. 100 keV passes without touching walls of solenoid then find current in the solenoid. Electron beam make angle ${30}^o$ with axis of solenoid.

• A. 0.4 A
• B. 0.8 A
• C. 1.2 A
• D. 1.6 A

Answer 1

The correct option is A 0.4 A

$\text{Given}:$ n=$2\times {10}^4$, d= 10 cm, K.E. = 100 keV, $\theta ={30}^o$

$\text{Solution}:$

Let v is the speed of the electron.

$K.E=\frac{1}{2}m{v}^2$

$\therefore v^2=\frac{2\times K.E}{m}$

Put the value into the formula

$v^2=\frac{2\times 100\times {10}^3\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}$

$⟹v=\sqrt{\frac{2\times 100\times {10}^3\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}}$= 187522892.37 m/s = $1.87\times {10}^8$m/s

let B is the magnetic field inside the solenoid We need to calculate the magnetic field Using formula of radius of path of the electron

$r=\frac{mvsin\theta }{Bq}$

$B=\frac{mvsin\theta }{qr}$

Put the value into the formula

$B=\frac{9.1\times {10}^{-31}\times 1.875\times {10}^8\sin 30}{1.6\times {10}^{-19}\times 5\times {10}^{-2}}$

B=0.0106 T

We need to calculate the current in the solenoid Using formula of magnetic field

$B={\mu }_0\times n\times I$

$I=\frac{B}{{\mu }_{0}n}$

Put the value into the formula

$I=\frac{0.0106}{4\pi \times {10}^{-7}\times 2\times {10}^4}$

I=0.421 A

$\text{Hence A is the correct option}$