A proton moving in a uniform magnetic field with v 1=1.19×10^6^m/s experiences f

Question

A proton moving in a uniform magnetic field with v 1=1.19×10^6^m/s experiences force F 1=1.86×10^16k^N . A second proton with v 2=2.43×10^6^m/s experiences F 2=4.20×10^16k^N in the same field.

Part A

What is the magnitude of B ?

Express your answer with the appropriate units.

Part B

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

Express your answer with the appropriate units.

  

A proton moving in a uniform magnetic field with v 1=1.19×10^6^m/s experiences force F 1=1.86×10^16k^N . A second proton with v 2=2.43×10^6^m/s experiences F 2=4.20×10^16k^N in the same field.

Part A

What is the magnitude of B ?

Express your answer with the appropriate units.

B =  

Part B

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

Express your answer with the appropriate units.

=

  

Explanation / Answer

We know that magnetic Force Fb = q v x B

1.86e-16 k = 1.6e-19*1.19e6i x By

By = 1.86e-16/(1.6e-19*1.19e6) = 0.000977 T

For second proton, - 4.2e-16 k = 1.6e-19*2.43e6j x Bx

Bx = - 4.2e-16/(1.6e-19*-2.43e6) = 0.00108 T

B = 0.00108 i +0.000977 j T

B = sqrt (0.00108^2 +0.000977 ^2)

= 0.00146 T answer

B) angle = arctan (0.000977 /0.00108)

= 42.13 degree above x axis

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