Okay, here it is in writing: A proton moving in a uniform magnetic field with v

Question

Okay, here it is in writing:

A proton moving in a uniform magnetic field with v 1=1.09×106^m/s experiences force F 1=1.54×1016k^N. A second proton with v 2=2.49×106^m/s experiences F 2=3.90×1016k^N in the same field.

What is the magnitude of B ?

What is the direction of B ? Give your answer as an angle measured ccw from the +x-axis.

That is everything. And it is velocity subscript 1, Force subscript 1, velocity subscript 2, Force subscript 2.

What is the magnitude of B? A proton moving in a uniform magnetic field with = 1.09 × 106 m/s experiences force F1 = 1.54 × 10-16 k N. A second proton with 2-249 × 106 J m/s experiences E,--3.90 × 10-16 k N in the same field Express your answer with the appropriate units BValueUnits Submit My Answers Give Up Part B What is the direction of B? Give your answer as an angle measured ccw from the +a-axis Express your answer with the appropriate units 8Value Units Submit My Answers Give Up

Explanation / Answer

Force is F = q*v*B*sin(theta)

theta is the angle between velocity and magnetic field

theta = 90 deg since the angle between i hat (along X-axis) and k hat (along z-axis)

then

F = q*v*B*sin(90)

q is the charge on proton = 1.6*10^-19 C

F = q*v*B

B = F/(q*v) = (1.54*10^-16)/(1.6*10^-19*1.09*10^6)

B = 8.83*10^-4 T along +y-axis

for second proton

B = F/(q*v) = (3.9*10^-16)/(1.6*10^-19*2.49*10^6)

B = 9.78*10^-4 T along +X-axis

then magnetic field is B = sqrt(Bx^2+By^2) = sqrt(9.78^2+8.83^2) = 13.17 T

direction of B is theta = tan^(-1)(8.83/9.78) = 42 deg CCW from +Xaxis

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