A proton moving with a velocity $2.5 imes 10^7$ m/s enters a magn...

Question

A proton moving with a velocity $2.5 	imes 10^7$ m/s enters a magnetic field of intensity 2.5 T at an angle of 30° with the magnetic field. Find the force on the proton. (Charge of proton $e = 1.6 	imes 10^{-19}$ C)

Accepted Answer

Magnetic force on a charged particle:
$$F = qvB\sin	heta$$
$$F = 1.6 	imes 10^{-19} 	imes 2.5 	imes 10^7 	imes 2.5 	imes \sin 30°$$
$$F = 1.6 	imes 10^{-19} 	imes 2.5 	imes 10^7 	imes 2.5 	imes 0.5$$
$$F = 1.6 	imes 10^{-19} 	imes 3.125 	imes 10^7$$
$$F = 5 	imes 10^{-12} 	ext{ N}$$

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