Question

An α-particle has a charge of +2e and a mass of 6.64 × 10-27 kg. It is accelerated from rest through a potential difference that has a value of 1.97 × 106 V and then enters a uniform magnetic field whose magnitude is 3.49 T. The α-particle moves perpendicular to the magnetic field at all times. What is (a) the speed of the α-particle, (b) the magnitude of the magnetic force on it, and (c) the radius of its circular path?

Answer 1

Answer with Explanation:

We are given that

Charge on alpha particle=q=2 e=$2\times 1.6\times 10^{-19} C$

1 e=$1.6\times 10^{-19} C$

Mass of alpha particle=m=$6.64\times 10^{-27} kg$

Potential difference,V=$1.97\times 10^6 V$

Magnetic field,B=3.49 T

a.Speed of alpha particle=v=$\sqrt{\frac{2 qV}{m}}$

By using the formula

$v=\sqrt{\frac{2\times 2\times 1.6\times 10^{-19}\times 1.97\times 10^6}{6.64\times 10^{-27}}$

$v=1.38\times 10^7 m/s$

b.Magnetic force,F$=qvB=2\times 1.6\times 10^{-19}\times 1.38\times 10^7\times 3.49=1.5\times 10^{-11} N$

F=$1.5\times 10^{-11} N$

c.Radius of circular path, r=$\frac{mv^2}{F}$

$r= \frac{6.64\times 10^{-27}\times (1.38\times 10^7)^2}{1.5\times 10^{-11}}$

r=0.084 m