Question

A particular solenoid (50 cm long with 2,000 turns of wire) has a current of 0.70 A and is in the void. An electron is fired at an angle of 10 from the axis of the solenoid from a point on the same axis. a) What should be the speed of the electron so that it just does not hit the inside of the solenoid of 16 cm in diameter? b) What will be the passage of the helical trajectory of the electron

Answer 1

(a) The magnitude of magnetic field inside the solenoid which is acting along the axis of the solenoid (50 cm long with 2,000 turns of wire) has a current of 0.70 A is B (say).

$B=\mu _{0}\left ( \frac{N}{l} \right )I=4\pi \times 10^{-7}\left ( \frac{2000}{0.5} \right )\times 0.70=3.518\ mT$

An electron is fired with a velocity of v at an angle of 10 from the axis of the solenoid from a point on the same axis. Then the component of velocity along the axis will be (v. cos 10) and perpendicular to it is (v. sin 10). It will make a helical path around the magnetic field direction. If it just does not hit the inside of the solenoid of 16 cm in diameter, then the radius of the helix (r) should be 8 cm. For a electron of charge e and mass m, the centrifugal force will be equal to the magnetic force.

$e\left ( v\sin 10^{o} \right )B=\frac{m\left ( v\sin 10^{o} \right )^{2}}{r}$

$\therefore v=\frac{eBr}{m\sin 10^{o}}=\frac{1.6\times 10^{-19}\times3.518\times 10^{-3} \times0.08}{9.1\times 10^{-31}\times \sin 10^{o}}=2.85\times 10^{8}\ m/s$

(b) The pitch of the helical path (p) can be written as

$p=T\left ( v\cos 10^{o} \right )=\frac{2\pi r}{eB}\left ( v\cos 10^{o} \right )$

$\therefore p=\frac{2\pi \times 0.08}{1.6\times 10^{-19}\times 3.518\times 10^{-3}}\left ( 2.85\times 10^{8}\times \cos 10^{o} \right )=4.42\times 10^{28}\ m$

The pitch is very high compared to the length of the solenoid. Therefore, it will eject from the solenoid from the circumference of the solenoid with rotating one full circle.