Chapter 26.1, Problem 1PP

To determine

The radius of the circular path if the speed of a proton is $7.5\times {10}^4\text{m/s}$ .

Answer to Problem 1PP

The radius of the circular path is $\boxed{r=9.79\times {10}^{-5}\text{m}}$ .

Explanation of Solution

Given:

Speed of the proton is $v=7.5\times {10}^4\text{m/s}$ .

Strength of magnetic field is $B=0.080\text{T}$

Formula used:

The expression for radius of the circular path is

  $r=\frac{mv}{Bq}$  ...... (1)

Here, $m,v,B$ and $q$ are mass of a particle, velocity, magnetic field strength and charge of a particle respectively.

Calculation:

Charge of a proton is $q=1.6\times {10}^{-16}\text{C}$ ,

Mass of a proton is $m=1.67\times {10}^{-27}\text{kg}$ ,

Substitute $7.5\times {10}^4\text{m/s}$ for $v$ , $0.080\text{T}$ for $B$ , $1.6\times {10}^{-16}$ for $q$ and $1.67\times {10}^{-27}\text{kg}$ for $m$ in equation (1),

  $\begin{array}{l}r=\frac{\left(1.67\times {10}^{-27}\right)\times \left(7.5\times {10}^4\right)}{\left(0.08\times \left(1.6\times {10}^{-19}\right)\right)}\\ r=\frac{12.53\times {10}^{-23}}{0.128\times {10}^{-19}}\\ r=97.89\times {10}^{-4}\text{m}\\ \boxed{r=9.79\times {10}^{-5}\text{m}}\end{array}$

Conclusion:

Hence, the radius of the circular path is $\boxed{r=9.79\times {10}^{-5}\text{m}}$ .