Question

A beam of singly ionized Lithium atoms is accelerated by a potential difference of 320 V. The beam passes through a magnetic
field of 0.015 T. What is the radius of curvature of the beam in the magnetic field?
0 0.50 m
O 0.25 m
O 0.74 m
0.46 m​

Answer 1

Given

m = Mass of lithium ions = $7m_p$

B = Magnetic field = 0.015 T

V = Potential difference = 320 V

To find

Radius of curvature of the beam = r

Solution

$m_p$ = Mass of proton = $1.67\times 10^{-27}\ \text{kg}$

Energy of beam

$K=qV$

Energy of particles

$K=\dfrac{1}{2}mv^2$

So

$\dfrac{1}{2}mv^2=qV\\\Rightarrow v=\sqrt{\dfrac{2qV}{m}}$

The magnetic force and centripetal force will balance each other

$qvB=\dfrac{mv^2}{r}\\\Rightarrow r=\dfrac{mv}{qB}\\\Rightarrow r=\dfrac{m\times \sqrt{\dfrac{2qV}{m}}}{qB}\\\Rightarrow r=\dfrac{7\times 1.67\times 10^{-27}\times \sqrt{\dfrac{2\times1.6\times 10^{-19}\times 320}{7\times 1.67\times10^{-27}}}}{1.6\times 10^{-19}\times 0.015}\\\Rightarrow r=0.45587\approx 0.46\ \text{m}$

The radius of the curvature of the beam in the magnetic field is $0.46\ \text{m}$