A magnetic field of magnitude 0.5 T points directly Into the paper. A proton wit

Question

A magnetic field of magnitude 0.5 T points directly Into the paper. A proton with a charge of +e traveling in the plane of the paper moves in a circle of diameter "2r" at a velocity of 1.234 m/s. Symbolically derive the equation for the Radius of the circle based on (1) the magnetic force law, and (2) the force required to cause Uniform circular motion, then use it to calculate the radius of the circle the proton moves in. You may use your book and or notes and/or the internet, ... (but no humans) to help you derive the equations.

Explanation / Answer

when the particle enter in the magnetic field magnetic force is applied

the magnetic force F = qVB

it will cause circular motion

the centripetel force F = mV2 / R

in the equilibrium condition

qVB = mV2 / R

so R = mV / qB

here m = 1.672 x 10-27 Kg V = 1.234 m/s q = 1.6 x 10-19 C B = 0.5 T

so the radius

r = ( 1.672 x 10-27 ) ( 1.234 ) / ( 1.6 x 10-19 ) (0.5)

r = 2.57 x 10-8 m Ans

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