Problem 9: A proton is moving with a velocity of v = vxi+vyj, where vx = 81 m/s

Question

Problem 9: A proton is moving with a velocity of v = vxi+vyj, where vx = 81 m/s and vy = 32 m/s. The proton enters a magnetic field B = Byj, where By = 2.3 T. Part (a) Express the magnetic force F in terms of the proton charge e, velocity v and the magnetic field B. Part (b) In this particular case, express the magnitude of the force, |F|, in in terms of e, vx, vy and By. Part (c) Calculate the numerical value of |F| in N. Part (d) What's the direction of the force? Use the right hand rule to find the direction of vector-product.

Explanation / Answer

a) F = e*(v cross B)

b) F = e*( (vxi + vyj) cross By j)

= e*vx*By k

c) F = 1.6*10^-19*81*2.3

= 2.98*10^-17 N

d) Positive z direction

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