A proton moves toward the top of the page with a speed of 4.77 middot 10^6 m/s i

Question

A proton moves toward the top of the page with a speed of 4.77 middot 10^6 m/s into a region of constant magnetic field as shown in the figure below. The magnitude of the magnetic field is 6.95 middot 10^-2 T and the direction of the field is out of the page. When the proton initially enters the magnetic field, the magnitude of the force on the proton (in N) is 1.275 times 10^-14 1.695 times 10^-14 2.255 times10^-14 2.999 times 10^-14 3.988 times 10^-14 5.304 times 10^-14 7.055 times 10^-14 9.383 times 10^-14 When the proton initially enters the magnetic field, the direction of the force on the proton is toward the top of the page toward the bottom of the page into the page. out of the page. to the right of the page to the left of the page

Explanation / Answer

1) F)5.304*10^-14 N

use, F = q*v*B*sin(theta)

= 1.6*10^-19*4.77*10^6*6.95*10^-2*sin(90)

= 5.304*10^-14 N

2) E) to the right of the page.

using right hand rule we determine the direction of magnetic force.

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