All Problems

Motion of Charged Particles in Electric and Magnetic Fields

Problem 3.378

A proton accelerated by a potential difference V=500kVV=500 \mathrm{kV} flies through a uniform transverse magnetic field with induction B=0.51B=0.51 T. The field occupies a region of space d=10 cmd=10 \mathrm{~cm} in thickness (Fig. 3.99). Find the angle α\alpha through which the proton deviates from the initial direction of its motion.

Reveal Answer
α=arcsin(dBq2mV)=30\alpha=\arcsin \left(d B \sqrt{\frac{q}{2 m V}}\right)=30^{\circ}