A proton moving with velocity v1 = 3.6 x 10^4 j (m/s) experiences a magnetic for

Question

A proton moving with velocity v1 = 3.6 x 10^4 j (m/s) experiences a magnetic force of 7.4 x 10^-16 i (N). A second proton moving on the x axis experiences a magnetic force of 2.8 x 10^-16 j (N).

Part A

Find the magnetic field.

Express your answer using two significant figures. Express your answer in terms of the variables i^, j^, and k^.

B = ? T

Part B

Find the velocity of the second proton.

Express your answer using two significant figures. Express your answer in terms of the variables i^, j^, and k^.

v2 = ? m/s

Explanation / Answer

B = F/qv = 7.4x10^-16 N/(1.60x10^-19 C)(3.6 x10^4 m/s) = .13 T
The direction of F must be in the direction of vxB. In this case, using the right hand rule, the direction of B is along the positive z-axis so
B = .13 k (T)

b)
B = F/qv
.13 T = 2.8x10^-16/(1.60x10^-19 C)v
v = 1.4x10^4 m/s

As above, use the right hand rule to get
v = -1.4x10^4 i (m/s)

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