A proton moves with v= 4x10^ 6 m/s and enters in B=1.9T and experiences a force...

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Question

A proton moves with v= 4x10^ 6 m/s and enters in B=1.9T and experiences a force of 8x 10^-9 n , calculate 0 between B and v

Answer 1

Answer #1

Given

proton moving with velocity v = 4*10^6 m/s

entered in to the magnetic field of B = 1.9 T and experiences a force of F = 8*10^-9 N

charge of wlwxtron is q = 1.6*10^-19 C

we knwo that the the force on a moving charge in a magnetic field is F = qv X B

F = q*v*B sin theta ,

solving for the angle theta ,

8*10^-9 = 1.6*10^-19*4*10^6*1.9 sin theta

theta = 41.14 degrees

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Answer 2

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