Chapter 19, Problem 42P

(a)

To determine

The speed of the proton as it leaves region 1 and enters region 2.

Answer to Problem 42P

The speed of the proton as it leaves region 1 and enters region 2 is $7.99\times {10}^5\text{m/s}$.

Explanation of Solution

The potential difference in region 1 is $3330\text{V}$. The magnetic field in region 2 and region 3 is $1.20\text{T}$. There is electric field in region 2.

As the proton passes out of region 1, the potential energy due to the potential difference is converted into kinetic energy.

    $e\Delta V=\frac{1}{2}m{v}^2$

Here, $e$ is the charge in proton, $\Delta V$ is the potential difference, $m$ is the mass, $v$ is the speed.

Re-write the above equation to get an expression for $v$.

    $v=\sqrt{\frac{2e\Delta V}{m}}$

Conclusion:

Substitute $1.602\times {10}^{-19}\text{C}$ for $e$, $3330\text{V}$ for $\Delta V$, $1.673\times {10}^{-27}\text{kg}$ for $m$.

    $\begin{array}{c}v=\sqrt{\frac{2\left(1.602\times {10}^{-19}\text{C}\right)\left(3330\text{V}\right)}{1.673\times {10}^{-27}\text{kg}}}\\ =7.99\times {10}^5\text{m/s}\end{array}$

The speed of the proton as it leaves region 1 and enters region 2 is $7.99\times {10}^5\text{m/s}$.

(b)

To determine

The magnitude and direction of the electric field in region 2 if the proton moves in straight line in region 2.

Answer to Problem 42P

The magnitude of the electric field is $9.58\times {10}^5\text{V/m}$ and the direction of the electric field is towards north.

Explanation of Solution

The potential difference in region 1 is $3330\text{V}$. The magnetic field in region 2 and region 3 is $1.20\text{T}$. There is electric field in region 2.

If the proton moves in straight line in region 2, the magnetic force is balanced by the electric force.

    $eE=evB$

Here, $e$ is the charge in proton, $E$ is the electric field, $B$ is the magnetic field, $v$ is the speed.

Re-write the above equation to get an expression for $E$.

    $E=vB$

The direction of the magnetic force is determined by right hand rule. The magnetic field is directed upward, the velocity of the proton is east ward. Thus by right hand rule the magnetic force will be towards south direction.

Conclusion:

Substitute $7.99\times {10}^5\text{m/s}$ for $v$, $1.20\text{T}$ for $B$.

    $\begin{array}{c}E=\left(7.99\times {10}^5\text{m/s}\right)\left(1.20\text{T}\right)\\ =9.58\times {10}^5\text{V/m}\end{array}$

To cancel the magnetic force, the electric field has to be in the opposite direction of the magnetic force. Thus the direction of the electric field is towards north.

The magnitude of the electric field is $9.58\times {10}^5\text{V/m}$ and the direction of the electric field is towards north.

(c)

To determine

The path of the proton in region 3.

Answer to Problem 42P

In region 3 the proton follows the path 2.

Explanation of Solution

The potential difference in region 1 is $3330\text{V}$. The magnetic field in region 2 and region 3 is $1.20\text{T}$. There is electric field in region 2.

The direction of the magnetic force is determined by right hand rule. The magnetic field is directed upward, the velocity of the proton is east ward. Thus by right hand rule the magnetic force will be towards south direction.

In region 3 there is only magnetic field. Thus the force on the proton is only the magnetic force. Thus the force on the proton is directed in south direction so it follows the path 2.

Conclusion:

In region 3 the proton follows the path 2.

(d)

To determine

The radius of the circular path in region 3.

Answer to Problem 42P

The radius of the circular path in region 3 is $6.95\text{mm}$.

Explanation of Solution

The potential difference in region 1 is $3330\text{V}$. The magnetic field in region 2 and region 3 is $1.20\text{T}$. There is electric field in region 2.

The magnetic force makes the proton move in circular path. Thus the magnetic force is equal to the centripetal force.

    $evB=\frac{m{v}^2}{r}$

Here, $e$ is the charge in proton, $B$ is the magnetic field, $m$ is the mass, $v$ is the speed, $r$ is the radius of the circular path.

Re-write the above equation to get an expression for $r$.

    $r=\frac{mv}{eB}$

Conclusion:

Substitute $1.602\times {10}^{-19}\text{C}$ for $e$, $1.20\text{T}$ for $B$, $1.673\times {10}^{-27}\text{kg}$ for $m$, $7.99\times {10}^5\text{m/s}$ for $v$.

    $\begin{array}{c}r=\frac{\left(1.673\times {10}^{-27}\text{kg}\right)\left(7.99\times {10}^5\text{m/s}\right)}{\left(1.602\times {10}^{-19}\text{C}\right)\left(1.20\text{T}\right)}\\ =6.95\text{mm}\end{array}$

The radius of the circular path in region 3 is $6.95\text{mm}$.