Magnetic Field Force of a proton

[PhysicsInNJ]

Homework Statement

A proton moves through a uniform magnetic field given by

= (10

−25.3(j-hat) + 30(k-hat))mT. At time t1, the proton has a velocity given by

= vx

+ vy(j-hat)+ (2.0 km/s)(k-hat) and the magnetic force on the proton is

B = (4.46 ✕ 10−17 N)

+ (1.76 ✕ 10−17 N)(j-hat).

At this instant, what is Vx?

Homework Equations

F= qV x B

The Attempt at a Solution

F/qB= V (using all the i-hat values since I am looking for Vx)
(4.46x10^-17)/(1.6x10^-19)(0.001)
This gives me= 278750 m/s

 

[gneill]

Your relevant equation involves a cross product, so each force component will involve a mix of contributions from other components of the velocity and field vectors.

I suggest that you expand the cross product in the equation symbolically first, then pick out useful equations from the force terms.

 

[PhysicsInNJ]

ok I have expanded the cross product and got this;

(Vy)(30) - (2000)(-25.6) = 30Vy + 512000
(2000)(10) - (Vx)(30) = 20,000 - 30Vx
(Vx)(-25.6) - (Vy)(-25.6) = -25.6Vx + 25.6Vy

So the relevant equation would be the second, since it has a Vx. I set that equal to 1.76E-17/ e and solved for Vx which did not work.

 

[gneill]

I don't see where you've accounted for the charge on the proton or the given units of the magnetic field terms (milli Teslas).

 

[PhysicsInNJ]

I accounted for the charge by dividing by e, and just redid the calculation with 0.03 instead of 30.

 

[gneill]

I accounted for the charge by dividing by e, and just redid the calculation with 0.03 instead of 30.

That accounts for the ##B\hat{k}## term. What about the ##B\hat{i}## term (10 mT)?

 

[PhysicsInNJ]

Thank you so much! that cleared it up!