If a weight of 500 gm is suspended at a point A and the length of rod is 10 m,find the force required at B to keep it stable and prevent rotation.
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Correct option is
A
0.5 T
Here, m=500g=0.5kg,I=4A,l=2.5m
As F=IlBsinθ
As the rod is suspended in air so,weight of thr rod is balanced by the magnetic force acting on the current carrying rod
mg=IlBsin90
o
, (∵θ=90
o
andF=mg)
∴B=
Il
mg
=
4×2.5
0.5×10
=0.5T
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