Chapter 20, Problem 72GP

(a)

To determine

The radius of curvature.

Answer to Problem 72GP

The radius of curvature is $r=4.14\times {10}^{-2}\text{m}$ .

Explanation of Solution

Given:

The mass of helium atom is $m=6.6\times {10}^{-27}\text{ kg}$ .

The charge of electron is $e=1.602\times {10}^{-19}\text{C}$

The voltage is $V=2400\text{ V}$ .

The magnetic field is $B=0.240\text{T}$ .

Formula used:

The force exerted $\left(F\right)$ on the charge moving with a velocity in a magnetic field is,

  $F=qvB$ …… (1)

Here, $q$ is the charge of charged particles, $v$ is the velocity of charged particle, and $B$ is the magnetic field.

The maximum force $\left(F_{\max }\right)$ is,

  $F_{\max }=ma$ …… (2)

Here, m is the mass and a is the acceleration.

The centripetal acceleration of charged particles ( a ) is,

  $a=\frac{v^2}{r}$ …… (3)

Calculation:

From the above three equations,

  $\begin{array}{l}{F}_{\max }=m\frac{v^2}{r}\\ qvB=\frac{m{v}^2}{r}\\ r=\frac{mv}{qB}\end{array}$

Substitute $\sqrt{\frac{2qV}{m}}$ for v .

  $\begin{array}{c}r=\frac{m\sqrt{\frac{2qV}{m}}}{qB}\\ =\sqrt{\frac{2mV}{q{B}^2}}\end{array}$

The charge of helium is,

  $\begin{array}{c}q=2e\\ =2\left(1.60\times {10}^{-19}\text{C}\right)\\ =3.2\times {10}^{-19}\text{C}\end{array}$

The radius of curvature is,

  $\begin{array}{c}r=\sqrt{\frac{2mV}{q{B}^2}}\\ =\sqrt{\frac{2\times 6.6\times {10}^{-27}\times 2400}{3.2\times {10}^{-19}C\times {\left(0.240\text{T}\right)}^2}}\\ =4.14\times {10}^{-2}\text{m}\end{array}$

Conclusion:

Thus, the radius of curvature is $r=4.14\times {10}^{-2}\text{m}$

(b)

To determine

The period of revolution.

Answer to Problem 72GP

The period of revolution is $T=5.39\times {10}^{-7}\text{s}$ .

Explanation of Solution

Given:

The mass of helium atom is $m=6.6\times {10}^{-27}\text{ kg}$ .

The charge of electron is $e=1.602\times {10}^{-19}\text{C}$

The voltage is $V=2400\text{ V}$ .

The magnetic field is $B=0.240\text{T}$ .

Formula used:

The time required for a particles of charge q moving with a constant speed v to make one circular revolution in a uniform magnetic field B is given by,

  $\begin{array}{c}T=\frac{2\pi r}{v}\\ =\frac{2\pi \sqrt{\frac{2mV}{q{B}^2}}}{\sqrt{\frac{2qV}{m}}}\\ =\frac{2\pi m}{B}\end{array}$

Calculation:

The period of revolution is,

  $\begin{array}{c}T=\frac{2\pi m}{qB}\\ =\frac{2\pi \left(6.6\times {10}^{-27}\right)}{3.2\times {10}^{-19}\times 0.240T}\\ =5.39\times {10}^{-7}\text{s}\end{array}$

Conclusion:

Thus, the period of revolution is $T=5.39\times {10}^{-7}\text{s}$ .