The force on a wire is a maximum of 4.39 N when placed between the pole faces of

Question

The force on a wire is a maximum of 4.39 N 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 11.2 cm, estimate the current in the wire if the field is 0.134T 2.93 times10^2 A If the wire is tipped so that it now makes an angle of 14.0 degree with the horizontal, what force will it now feel? 4.26N Two variables changed due to the tipping, 1. The angle between the current and the magnetic is no longer 90 deg, 2. The length of the wire in the magnetic field also changed, Use trigonometry to find the length of the wire in the magnetic field.

Explanation / Answer

a) We find the current from the length of wire in field: F = ILB;

4.39N = I(0.112m)(0.134T), which gives I =  2.93 x 102A.

(b) The new force is F = ILBsinq = Fsinq = (4.39N)sin76° =  4.26 N.

               Note that the wire could be tipped either way.

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