Chapter 26.1, Problem 12SSC

To determine

The radius of the circular path taken by the proton when it moves at a speed of $8.4\times {10}^4{\text{ms}}^{\text{-1}}$ in a magnetic field of $12.0\text{mT}$ .

Answer to Problem 12SSC

The radius of the circular path is 0.07 m.

Explanation of Solution

Given:

The speed of a proton in a circular path is $v=8.4\times {10}^4\text{m}\cdot {\text{s}}^{\text{-1}}$ .

The strength of a magnetic field is $B=12.0\text{mT}$ .

Formula used:

Expression of the radius of the circular path followed by an ion in the spectrometer is

  $r=\frac{mv}{qB}$…… (1)

Calculation:

Substitute $8.4\times {10}^4{\text{ms}}^{\text{-1}}$ for $v$ , $1.67\times {10}^{-27}\text{kg}$ for $m$ , $1.6\times {10}^{-19}\text{C}$ for $q$ and 12.0 $\text{mT}$ for $B$ in equation (1)

  $r=\frac{\left(1.67\times {10}^{-27}\text{kg}\right)\times \left(8.4\times {10}^4\text{m}\cdot {\text{s}}^{\text{-1}}\right)}{\left(1.6\times {10}^{-19}\text{C}\right)\times \left(12.0\text{mT}\right)}$

  $\begin{array}{l}r=\frac{\left(1.67\times {10}^{-27}\text{kg}\right)\times \left(8.4\times {10}^4\text{m}\cdot {\text{s}}^{\text{-1}}\right)}{\left(1.6\times {10}^{-19}\text{C}\right)\times \left(12.0\times {10}^{-3}\text{T}\right)}\\ \boxed{r=0.07\text{m}}\end{array}$

Conclusion:

Hence, the radius of the circular path is 0.07 m.