A proton (mass m = 1.67 times 10^{-27} , kg a | vedclass

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Question

(A) The magnetic force acting on a charged particle moving in a magnetic field is given by $F = qvB \sin(	heta)$.Since the proton enters perpendicula

Vedclass · Accepted Answer

$6.5 	imes 10^{15} \, m/s^2$

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(A) The magnetic force acting on a charged particle moving in a magnetic field is given by $F = qvB \sin(	heta)$.Since the proton enters perpendicular to the magnetic field,$	heta = 90^\circ$,so $\sin(90^\circ) = 1$.Thus,the force is $F = qvB$.According to Newton's second law,$F = ma$,therefore $ma = qvB$.The acceleration $a$ is given by $a = \frac{qvB}{m}$.Substituting the given values:$a = \frac{(1.6 	imes 10^{-19} \, C) 	imes (3.4 	imes 10^7 \, m/s) 	imes (2 \, Wb/m^2)}{1.67 	imes 10^{-27} \, kg}$.$a = \frac{10.88 	imes 10^{-12}}{1.67 	imes 10^{-27}} \, m/s^2$.$a \approx 6.515 	imes 10^{15} \, m/s^2$.Rounding to the given options,$a = 6.5 	imes 10^{15} \, m/s^2$.

$A$ proton (mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$) enters perpendicular to a magnetic field of intensity $B = 2 \, Wb/m^2$ with a velocity $v = 3.4 \times 10^7 \, m/s$. The acceleration of the proton is:

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