Chapter 20, Problem 25P

(a)

To determine

The angle of leaving.

Answer to Problem 25P

The angle at which proton leaves is ${45}^{\circ }$ .

Explanation of Solution

Given:

The proton enters from a field-free region into the uniform magnetic field and then leaves in the same.

The angle at which proton enters is $45°$ with a velocity of $2\times {10}^5\text{ m/s}$ .

Magnetic field is 0.850 T.

  

Calculation:

Consider the figure shown below.

  

Since a proton is making a curve which when extended forms a circle and the distance $x$ is the chord of this circle. Now, if perpendicular bisector is drawn using this chord, then it will bisect the central angle of the circle, which proves angle ${\theta }_4$ is equal to the angle ${\theta }_5$ . Using the geometry concepts, angle ${\theta }_1$ is equal to the angle ${\theta }_2$ due to vertically opposite angle. Angle ${\theta }_2$ is equal to the angle ${\theta }_4$ due to complementary angles of ${\theta }_3$ . Similarly, angle ${\theta }_7$ is equal to the angle ${\theta }_5$ due to complementary angles of ${\theta }_6$ . And, angle ${\theta }_7$ is equal to the angle ${\theta }_8$ due to vertically opposite angle. This proves that the entering angle ${\theta }_1$ is equal to the exit angle ${\theta }_8$ which is equal to the value of $45°$

Conclusion:

The angle at which proton leaves is $45°$ .

(b)

To determine

The distance of proton to exit the field.

Answer to Problem 25P

The distance $x$ at which proton exits the field is $3.5\times {10}^{-3} m$

Explanation of Solution

Given:

The proton enters from a field-free region into the uniform magnetic field and then leaves in the same.

The angle at which proton enters is $45°$ with a velocity of $2\times {10}^5\text{ m/s}$ .

Magnetic field is 0.850 T.

  

Formula Used:

The radius of the particle in the magnetic field is calculated as $r=\frac{mv}{qB}$ .

Calculation:

The distance $x$ is equal to the product of cosine angle and the radius of the circle that is

  $\begin{array}{l}x=2r\cos \theta =2\frac{mv}{qB}\cos \theta \\ x=2\frac{\text{(1}{\text{.67×10}}^{\text{-27}}{\text{kg)(2×10}}^{\text{5}}\text{m/s)}}{\text{(1}{\text{.6×10}}^{\text{-19}}\text{C)(0}\text{.85T)}}\cos 45°\\ x=3.5\times {10}^{-3}\text{m}\end{array}$

Conclusion:

The distance $x$ at which proton exits the field is $3.5\times {10}^{-3} \text{m}$