Chapter 26, Problem 6STP

To determine

To Choose:The correct option.

Answer to Problem 6STP

The correct option is (C).

Explanation of Solution

Given:

Radius of the circular path is $r=0.52\text{m}$ .

The strength of the magnetic field is $B=0.45\text{T}$

The mass of the individual proton is $m=1.67\times {10}^{-27}\text{kg}$

Formula used:

The expression for magnetic force on the proton is

  $F=qvB\sin \theta$  ...... (1)

Here, $q,v,B$ and $\theta$ are the charge of proton, speed of proton, strength of the magnetic field and angle between the direction of velocity and direction of magnetic field respectively.

Calculation:

Since the magnetic field is perpendicular to the velocity of the beam, then the angle between velocity and the field should automatically be $90°$ .

Substitute $90°$ for $\theta$ in equation (1)

  $F=qvB\sin 90°$

  $F=qvB$  ...... (2)

When a proton beam is moving in a circular path, then the centripetal force can be expressed as

  $F=\frac{m{v}^2}{r}$  ...... (3)

Here, $m$ is the mass of proton, $v$ is the speed of the proton and $r$ is the radius of the circular path.

Since the necessary centripetal force is provided by the magnetic field, therefore,

  $qvB=\frac{m{v}^2}{r}$

  $qB=\frac{mv}{r}$  ...... (4)

Rearranging the equation for speed $v$ from equation (1),

  $v=\frac{qBr}{m}$..................... (5)

Substitute $1.6\times {10}^{-19}\text{C}$ for $q$ , $0.45\text{T}$ for $B$ , $0.52\text{m}$ for $r$ and $1.67\times {10}^{-27}\text{kg}$ for $m$ in equation (4)

  $\begin{array}{l}v=\frac{\left(1.6\times {10}^{-19}\text{C}\right)\times \left(0.45\text{T}\right)\times \left(0.52\text{m}\right)}{\left(1.67\times {10}^{-27}\text{kg}\right)}\\ \boxed{v=2.24\times {10}^7\text{m/s}}\end{array}$

Conclusion:

Hence, the required speed of the proton is $\boxed{v=2.24\times {10}^7\text{m/s}}$ .