Chapter 20, Problem 67GP

To determine

The magnitude and direction of the magnetic field.

Answer to Problem 67GP

The magnitude of the magnetic field is s $B=3\text{ T}$ .

The direction of the magnetic field must point upwards.

Explanation of Solution

Given:

The diameter of circular path is $\text{2}\text{.0 km}$ .

The momentum is $\rho =4.8\times {10}^{-16}\text{ kg}\cdot \text{m}/\text{s}$ .

Formula used:

The maximum force exerted $\left(F_{\max }\right)$ on the charge moving with a velocity in a magnetic field is,

  $F_{\max }=qvB$ .... (1)

Here, $q$ is the charge of charged particles, $v$ is the velocity of charged particle, and $B$ is the magnetic fields.

Calculation:

The expression for the maximum force of $\left(F_{\max }\right)$ is,

  $F_{\max }=ma$

Here, $m$ is the mass of the charge and $a$ is the acceleration.

Substitute $\frac{v^2}{r}$ for a .

  $\begin{array}{c}{F}_{\max }=m\left(\frac{v^2}{r}\right)\\ =\frac{m{v}^2}{r}\end{array}$ .... (2)

Equating the Equation (1) and (2).

  $\begin{array}{l}qvB=\frac{m{v}^2}{r}\\ mv=qBr\end{array}$

Here, r is radius of curvature and q is charge of charged particles.

The momentum $\left(\rho \right)$ is,

  $\begin{array}{c}\rho =mv\\ =qBr\\ B=\frac{\rho }{qr}\end{array}$

The radius of the circular path is,

  $\begin{array}{c}r=\frac{d}{2}\\ =\frac{2}{2}\\ =1\text{km}\times \left(\frac{{10}^3\text{m}}{1\text{km}}\right)\\ =1\times {10}^3\text{m}\end{array}$

The magnitude of the magnetic field is,

  $\begin{array}{c}B=\frac{\rho }{qr}\\ =\frac{\left(4.8\times {10}^{-16}\text{kg}\cdot \text{m}/\text{s}\right)}{\left(1.60\times {10}^{-19}\right)\times 1.0\times {10}^3\text{m}}\\ =3.0\text{T}\end{array}$

The direction of the magnetic field must point upward to cause centripetal force that sheers the proton clockwise in circular path.

Conclusion:

Thus, the magnitude of the magnetic field is $B=3\text{ T}$ and the direction of the magnetic field must point upward