4.a) A proton enters a uniform magnetic field B in a perpendicular direction. Th

Question

4.a) A proton enters a uniform magnetic field B in a perpendicular direction. The proton is moving out page with a speed of 1.00 x 10^7 m/s when it enters the magnetic field After entering the magnetic field the proton experiences acceleration of 2.00 x 10^13 m/s^2 in + x direction. Determine the magnitude and direction of the magnetic field B? b) Find the direction of the magnetic field acting on an electron moving through two situations shown in the figure below. (Note: The directions of velocity and force are shown in each case.)

Explanation / Answer

The magnetic force is perpendicular to the plane formed by the direction of velocity, V, and magnetic field, B. The direction of that perpendicular is found by the right hand rule; "curl fingers of right hand from V toward B and thumb points in direction of force (opposite that for negative charges).
The vector equation is the cross product;

F = qV X B= ma

The magnitude of the force is;

F = qVBSin(a) , where , a , is the angle between V & B.

So in your problem;

F = (1.6 x 10^-19)(10^7)(B)Sin(90)

= ma = 9.1905*10^-28 kg * 2*10^13

B= 9.1905*2 *10^-15/ 1.6*10^-12 =0.0114T

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