In a typical experiment to study the force between electric currents in identica

Question

In a typical experiment to study the force between electric currents in identical fixed and movable wires, were seperated 0.687 cm and the length of the movable wire was 26.5cm. The apparatus was set up so as to produce as attractive force when a current of 9.3A passed through the wires in the same directions. Assume the magnetic permeability of a free space Uo= 4piE-7 T.m/A. what mass(in milligram) would have to be removed from the weight pan on the movable wire to yield the attractive magnetic force when the given current passes through the fixed wire?

Explanation / Answer

the magnetic force is

F = BIL

where B is magnetic field,I is current and L is length of conductor

the weight is

w = mg

where m is mass and g = 9.8 m/s^2

Here,F = w

or m = (BIL/g)

where I = 9.3 A and L = 26.5 cm = 26.5 * 10^-2 m

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