The force on a wire is a maximum of 7.00×102Nwhen placed between the pole faces

Question

The force on a wire is a maximum of 7.00×102Nwhen placed between the pole faces of a magnet. The current flows horizontally to the right and the magnetic field is vertical. The wire is observed to "jump" toward the observer when the current is turned on.

I have found A, which is I= 1.6 A. I can't figure out how to get the right answer for B.

A.) If the pole faces have a diameter of 17.0 cm , estimate the current in the wire if the field is 0.25 T.

B.) If the wire is tipped so that it makes an angle of 19.0 with the horizontal, what force will it now feel?

Explanation / Answer

Let us use left-hand rule for magnetic fields :

The wire would only jump towards the observer if the field is going from bottom to top ie
A) Top is the South pole face

F = BIL (B->flux; I->current, L->length of wire in the field)
I = F/BL
I = 7.00x10^-2/0.25 x 0.17
I = 1.647 A

B) F(new) =F(old) cos 19
= 7.00x10^-2 x cos19
= 6.618x10^-2 N

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