Chapter 26, Problem 42A

To determine

The strength of the electron experience if its path is a circle of radius 5.0 cm.

Answer to Problem 42A

  $\boxed{B=4.53\text{mT}}$ .

Explanation of Solution

Given:

The potential difference is $V=4.5\text{kV}$ .

Radius of the circular path is $r=5.0\text{cm}$ .

Formula used:

The charge to mass ratio of an ion in a mass spectrometer is given by

  $\frac{q}{m}=\frac{2V}{B^2{r}^2}$  ...... (1)

Here, $m,v,q$ and $B$ and $r$ are mass, velocity, charge and magnetic field and radius respectively.

Calculation:

Simplified equation (1) for $B$ ,

  $B=\frac{1}{r}\sqrt{\frac{2mV}{q}}$  ...... (2)

Substitute $5.0\text{cm}$ for $r$ , $4.5\text{kV}$ for $V$ , $9.11\times {10}^{-31}\text{kg}$ for $m$ and $1.60\times {10}^{-19}\text{C}$ for $q$ in equation (2)

  $\begin{array}{l}B=\frac{1}{r}\sqrt{\frac{2mV}{q}}\\ B=\frac{1}{\left(5.0\text{cm}\right)}\times \sqrt{\frac{2\times 9.11\times {10}^{-31}\text{kg}\times 4.5\text{kV}}{1.60\times {10}^{-19}\text{C}}}\\ B=\frac{1}{\left(5.0\times {10}^{-2}\text{m}\right)}\times \sqrt{\frac{2\times \left(9.11\times {10}^{-31}\text{kg}\right)\times \left(4.5\times {10}^3\text{V}\right)}{\left(1.60\times {10}^{-19}\text{C}\right)}}\\ \boxed{B=4.53\text{mT}}\end{array}$

Conclusion:

Hence, the strength of the magnetic field is $\boxed{B=4.53\text{mT}}$ .