proton moves through a uniform magnetic field given by B = (10 ihat ? 16.7 jhat

Question

proton moves through a uniform magnetic field given by B = (10 ihat ? 16.7 jhat + 30 khat) mT. At time t1, the proton has a velocity given by v = v x ihat + vy jhat + (2000 m/s) khat and the magnetic force on the proton is F = (3.88 10-17 N) ihat + (2.31 10-17 N) jhat. (a) At that instant, what is vx? (b) What is vy?

Explanation / Answer

F=q(E+VxB) F,V,B,E are vectors since, E=0 F=q(VxB) ; B = (10i + 16.7j + 30k) mT ; F = (3.88X10^-17)i + (2.31x10^-17 )j N V = Vxi + Vyj + 2000k (m/s) ; q = 1.67x10^-19 C => (3.88X10^-17)i + (2.31x10^-17 )j = 1.67x10^-19 [(Vxi + Vyj + 2000k)x(10i + 16.7j + 30k)x10^-3] => (3.88X10^-17)i + (2.31x10^-17 )j = 1.67x10^-22 [16.7Vxi - 30Vxj - 10Vyk + 30Vyi + 20000j - 2000x16.7i] compare i and j components to get value of Vx and Vy

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