A proton is accelerated to a speed of 4.38E+6 m/s in a horizontal direction alon

Question

A proton is accelerated to a speed of 4.38E+6 m/s in a horizontal direction along the x-axis. It then enters a magnetic field of 4.48 T pointing into the screen (the y-direction).

(a) Calculate the radius of the circular path of the particle.  

(b) Find the electric field vector that would cause the proton to move in a straight line.(Note: Use the coordinate system where the x-direction points in the direction of the proton's initial motion, the y-direction points in the direction of the magnetic fields, and the z-direction is perpendicular to both of these[see figure].) The components of the electric field vector are

Ex =  
Ey =  
Ez =  

Explanation / Answer

Given:-

Proton speed (v) = 4.38

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