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Question

(A) The force $F$ on a charge $q$ moving with velocity $v$ in a magnetic field $B$ is given by $F = qvB \sin 	heta$. Since the proton moves perpendic

Vedclass · Accepted Answer

$8 	imes 10^{-12}$

Vedclass · Answer

(A) The force $F$ on a charge $q$ moving with velocity $v$ in a magnetic field $B$ is given by $F = qvB \sin 	heta$. Since the proton moves perpendicular to the field,$	heta = 90^{\circ}$,so $F = qvB$.Given kinetic energy $K = 2 \ MeV = 2 	imes 10^6 	imes 1.6 	imes 10^{-19} \ J = 3.2 	imes 10^{-13} \ J$.Using $K = \frac{1}{2}mv^2$,the velocity $v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 	imes 3.2 	imes 10^{-13}}{1.6 	imes 10^{-27}}} = \sqrt{4 	imes 10^{14}} = 2 	imes 10^7 \ m/s$.Now,$F = (1.6 	imes 10^{-19} \ C) 	imes (2 	imes 10^7 \ m/s) 	imes (2.5 \ T)$.$F = 1.6 	imes 10^{-19} 	imes 5 	imes 10^7 = 8 	imes 10^{-12} \ N$.

$A$ proton is moving perpendicular to a uniform magnetic field of $2.5 \ T$ with $2 \ MeV$ kinetic energy. The force on the proton is . . . . . . $N$. (Mass of proton $= 1.6 \times 10^{-27} \ kg$,charge of proton $= 1.6 \times 10^{-19} \ C$)

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