Chapter 26, Problem 42P

To determine

The magnetic intensity.

Answer to Problem 42P

  $3.3T$

Explanation of Solution

Given:

Kinetic energy is $998GeV$

The proton’s rest mass is $0.938GeV/{\text{c}}^2$

Mass of proton is $1.67\times {10}^{-27}kg$

Radius is $1.0km=1.0\times {10}^{3}m$

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:

Mass is constant during the revolution.

The total energy of an electron is

  $E=m{c}^2=KE+m_0{c}^2=998GeV+0.938GeV=999GeV$

So, the mass is $\frac{999GeV}{c^2}$

Calculate the speed form $m=\frac{m_0}{{\left[1-\left(\frac{v^2}{c^2}\right)\right]}^{\frac{1}{2}}}$

  $\frac{999GeV}{c^2}=\frac{\left(\frac{0.938GeV}{c^2}\right)}{{\left[1-\left(\frac{v^2}{c^2}\right)\right]}^{\frac{1}{2}}}$

This gives ${\left[1-\left(\frac{v^2}{c^2}\right)\right]}^{\frac{1}{2}}=9.39\times {10}^{-4}$

  $v=1.00c$

By using the formula $B=\frac{mv}{qr}=\frac{m_{0}v}{qr{\left[1-\left(\frac{V^2}{c^2}\right)\right]}^{\frac{1}{2}}}$

  $B=\frac{\left(1.67\times {10}^{-27}kg\right)\left(3.00\times {10}^{8}m/s\right)}{\left(1.6\times {10}^{-19}C\right)\left(1.0\times {10}^{3}m\right)\left(9.39\times {10}^{-4}\right)}=3.3T$

Conclusion: The required value is $3.3T$ .