1. [1pt] A straight wire 0.010 m long lies in the field of several horseshoe mag

Question

1. [1pt]
A straight wire 0.010 m long lies in the field of several horseshoe magnets. (The horseshoe magnets are lying side by side, with their magnetic fields pointing in the same direction; see diagram `Current balance and circuit' in the lab manual.) The wire is perpendicular to the magnetic field, B. The magnitude of B is uniform along the length of the wire and has the value 0.34 T. The wire carries a current, I, of 2.2 A. What is the magnitude of the force exerted on the wire by the magnetic field?

Explanation / Answer

Magnetic force on the current carrying wire

F = I B L sin90

    = ( 2.2 A) ( 0.34 T ) (0.01 m)

    =0.00748 N

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