A proton moving with velocity v 1=3.6×104 j ^(m/s) experiences a magnetic force

Question

A proton moving with velocity v 1=3.6×104j^(m/s) experiences a magnetic force of 7.4×1016i^(N). A second proton moving on the x axis experiences a magnetic force of 2.8×1016j^(N).

A) Find the magnetic field.

Express your answer using two significant figures. Express your answer in terms of the variables i^, j^, and k^. Note that you need to enter the "hat", not just the letter!

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^. Note that you need to enter the "hat", not just the letter!

Explanation / Answer

  a)
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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