Answer to Question #127346 in Electricity and Magnetism for Reeshabh Kumar

Question

protons of k.e 10^12 are injected into a uniform magnetic field of strength 10 Tesla .A magnetic field exists only inside the cylindrical region of diameter 50 cm and is parallel to the axis of cylinder at the point of injection. The proton beam is directed towards the axis of the cylinder and is perpendicular to it . By the time the beam exists the magnetic field changes direction by

Answer 1

We can find the radius of the 'piece' of trajectory using Newton's second law, Lorentz, and centripetal force:

$$qvB=\frac{mv^2}{R},\\\space\\ R=\frac{mv}{qB}.$$

The speed of a single proton is

$$v=\sqrt{\frac{2E_K}{m}}.$$

Hence:

$$R=\frac{\sqrt{2mE_K}}{qB}=\\\space\\ =\frac{\sqrt{2(1.67\cdot10^{-27})10^{12}}}{1.602\cdot10^{-19}\cdot10}=45\text{ m}.$$

Now it's time to draw the situation:

Our goal for now is to find angle θ.

$$\alpha=\text{ atan}\frac{r}{R}=\text{ atan}\frac{0.5}{45.7}=0.63°.$$

We know that the sum of interior angles of trapezoid ABCD is 360°. Therefore:

$$(\theta+90)=360-0.63\cdot2-90-90,\\ \theta=88.7°.$$

Thus, the deflection d is

$$d=r\text{ cos}\theta=0.011\text{ m},$$

or 1.1 cm. The beam changes its direction for $90°-\theta=1.3°.$