A long wire carrying a 6.25A current reverses direction by means of two right-an
ID: 1283582 • Letter: A
Question
A long wire carrying a 6.25A current reverses direction by means of two right-angle bends, as shown in the figure (Figure 1) . The part of the wire where the bend occurs is in a magnetic field of 0.666T confined to the circular region of diameter 75.0cm , as shown. Assume that I= 6.25A and d=49.0cm .
Find the magnitude of the net force that the magnetic field exerts on this wire.
A long wire carrying a 6.25A current reverses direction by means of two right-angle bends, as shown in the figure (Figure 1) . The part of the wire where the bend occurs is in a magnetic field of 0.666T confined to the circular region of diameter 75.0cm , as shown. Assume that I= 6.25A and d=49.0cm . Find the magnitude of the net force that the magnetic field exerts on this wire.Explanation / Answer
As current in the two parallel lines is oppositeto each other therefore the force acting on them will be oppositeto each other. Therefore net force exerting on these two parallelwires is zero.
So net force on the entire wire will be equal to forced achingon 45 cm wire .
Net force on the wire F = Bil
magnetic field ( B ) = 0.666T
current through the wire ( I ) = 6.25 A
length of the wire ( l ) = 49 cm = 49 * 10-2m
F = B I l
= ( 0.666 T ) ( 6.25 A ) ( 49 *10-2 m )
= 2.04
I hope it helps you
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