Given is:-
Length of the wire
Mass of the wire
Magnetic field
Current
Now,
part-a
According to Newton’s second law, the weight of the wire is balanced by the vertical component of the tension force, therefore
eq-1
and the magnetic force is balanced by the horizontal component of the tension force
eq-2
we know that
Hence dividing eq-2 by eq-1 we get
or
by plugging all the values we get
which gives us
Part-b
From eq-1 we get
by plugging all the values we get
which gives us
A horizontal wire is hung frem the celing of a room by two massless strings. The...
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ignment FULL SCREEN PRINTER VERSION BACK NEXT Chapter 21, Problem 40 GO Chalkboard Video A horizontal wire is hung from the ceiling of a room by two massless strings. The wire has a length of 0.10 m and a mass of 0.005 kg. A uniform magnetic field of magnitude 0.055 T is directed from the ceiling to the floor. When a current of I - 42 A exists in the wire, the wire swings upward and, at equilibrium, makes an...
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