A proton moves to the left with speed of 6.34-106 m/s into a region magnetic fie

Question

A proton moves to the left with speed of 6.34-106 m/s into a region magnetic field as shown the magnitude of the magnetic field is 2.54. 10-2 T and the direction of the field is into the page. x x x x x x x x x x x x x x x x x x x x When the proton initially enters the magnitude field the magnitude of the force on the proton (in N) is A:2.577times B2.912times 10-14 C:3.290 times 10-14 E. 201 times 10-14 F:747 times 10-14 G: 5.364 times 10-4 H:6.062 times 10-14 When the proton initially enters the magnitude field the direction of the force on the proton When the proton enters the magnitude field the direction of the force the proton is A:toward the top of the B:toward the bothtom of the C:into the page D:out of the page E: to the right of the page F: to the left of the page

Explanation / Answer

Sorry it was near deadline to answer the question.... 1. F=Bqv, just insert numbers, where B is the magnitude of the magnetic field, q of proton is the same as a standard 'e', and v is the velocity. so A. 2. use the right hand rule. Make your right hand point at the direction of the magnetic field first, then the velocity of the proton. The direction of the your thumb indicates the force. So A.

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