The force on a wire is a maximum of 4.00 x 10^-2N when placed between the pole f

Question

The force on a wire is a maximum of 4.00 x 10^-2N when 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. If the pole faces have a diameter of 17.0cm , a)estimate the current in the wire in A if the field is 0.26T.b)If the wire is tipped so that it makes an angle of 28.0 degree with the horizontal, what force will it now feel in N?

Explanation / Answer

Using the left-hand rule for magnetic fields (see sources) : 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 = 4.00 x 10^-2/(0.26 x 0.17 ) I = 0.90497 A C) F(new) =F(old) cos 10 = 4.00 x 10^-2 x cos28 = 0.035 N

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