A proton that has a speed equal to 120 106 ms enters a regio

A proton that has a speed equal to 1.20 106 m/s enters a region with a uniform magnetic field that has a magnitude of 0.680 T and points into the page, as shown in the figure below. The proton enters the region at an angle = 60°.Find the exit angle and the distance d.

exit angle °

distance mm

Solution

Because of the symmetry of the circle the two intersecting points have the chord and tangent at same angle

= 60 deg

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theta = 180 - 90 - 60 = 30 deg

R / D = cos 30 = 0.866

D = R / 0.866

R = m v / q B

= (1.67 * 10^-27 * 1.20 * 10⁶) / (1.6 * 10^-19 * 0.680)

= 0.0184 m

D = R / 0.866 = 0.0184 / 0.866 = 0.0213 m

= 21.3 mm