Chapter 26, Problem 36A

To determine

To find: The speed of a proton moving in present of magnetic field.

Answer to Problem 36A

The resultant speed of the proton is $6.89\times {10}^5\text{m/s}$ .

Explanation of Solution

Given:

The radius of the path $0.2\text{m}$

The magnetic field strength is $0.036\text{T}$

Mass of a proton is $\text{1}\text{.67}\times {\text{10}}^{-27}\text{kg}$

Formula:

The force experience by an electron in present of magnetic field

  $F=qvB$  ...... (1)

Here, $v$ is the velocity of electron, $F$ is the force, $q$ is the charge of ions and $B$ is the magnetic field strength.

Centripetal force is given by

  $\text{F=}\frac{m{v}^2}{r}$  ...... (2)

Here, $m$ is the mass of the proton, $v$ is the circular velocity of the particle and $r$ is the radius of curvature.

From (1) and (2)

  $qvB\text{=}\frac{m{v}^2}{r}$

  $r=\frac{mv}{qB}$  ...... (3)

Calculation:

Substitute $0.2\text{m}$ for radius $\left(r\right)$ , $0.036\text{T}$ for magnetic field $(B)$ and $1.67\times {10}^{-27}\text{kg}$ for mass ( $m$ ) in equation (3)

  $\begin{array}{c}r=\frac{mv}{qB}\\ v=\frac{rqB}{m}\\ =\frac{0.2\times 1.6\times {10}^{-19}\times 0.036}{1.67\times {10}^{-27}}\\ =6.89\times {10}^5\text{m/s}\end{array}$

Conclusion:

Hence, the resultant speed of the proton is $6.89\times {10}^5\text{m/s}$ .