Chapter 26, Problem 54GP

To determine

The radius of curvature of path.

Answer to Problem 54GP

  $2.2mm$

Explanation of Solution

Given:

The rest mass of electron is $9.11\times {10}^{-31}kg$

Magnetic field is $1.8T$

Speed is $0.92c$

Charges on an electron is $1.6\times {10}^{-19}C$

Formula used:

The magnetic force provides the radial acceleration:

  $qvB=\frac{m{v}^2}{r}$ or $B=\frac{mv}{qr}=\frac{m_{0}v}{qr{\left[1-\left(\frac{V^2}{c^2}\right)\right]}^{\frac{1}{2}}}$

Where,

  $m_0$ is rest mass

  $m$ is relativistic mass

  $v$ is velocity

  $r$ is radius

  $q$ is charges

  $B$ is magnetic force

  $c$ is speed of light, $3.00\times {10}^{8}m/s$

Calculation:

The magnetic force provides the radial acceleration

  $r=\frac{m_{0}v}{qB{\left[1-\left(\frac{v^2}{c^2}\right)\right]}^{\frac{1}{2}}}$

  $r=\frac{\left(9.11\times {10}^{-31}kg\right)\left(0.92\right)\left(3.00\times {10}^{8}m/s\right)}{\left(1.6\times {10}^{-19}C\right)\left(1.8T\right){\left[1-{\left(0.92\right)}^2\right]}^{\frac{1}{2}}}$

  $2.2\times {10}^{-3}m=2.2mm$

Conclusion: The required value is $2.2mm$ .