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A square loop of length L and resistence R is pulled out of a magnetic field B . Which is perpendicular to the loop slowly and uniformly in t(sec).One edge of the loop was initially placed at the boundary of magnetic field.Find the work done in pulling the loop out: a)B 2 L 4 /Rt b)B 2 L 3 /Rt c)B 2 L2/Rt d)BL 3 /Rt

   A square loop of length L and resistence R is pulled out of a magnetic field B . Which is perpendicular to the loop slowly and uniformly in t(sec).One edge of the loop was initially placed at the boundary of magnetic field.Find the work done in pulling the loop out:
a)B2L4/Rt          b)B2L3/Rt
c)B2L2/Rt                 d)BL3/Rt  

Grade:12

1 Answers

ROSHAN MUJEEB
askIITians Faculty 829 Points
one year ago
Speed of the loop should be
v=lt=0.52=0.25m/sv=lt=0.52=0.25m/s
Induced emf,eBvl=(1.0)(1.0)(0.25)(0.5)eBvl=(1.0)(1.0)(0.25)(0.5)
=0.125V=0.125V
∴∴Current in the loopi=eR=0.12510i=eR=0.12510
=1.25×10−2A=1.25×10-2AThe magnetic force on the left arm due to the magnetic field is
Fm=ilB=(1.25×10−2(0.5)(1.0)Fm=ilB=(1.25×10-2(0.5)(1.0)
=6.25×10−3N=6.25×10-3N
to pull the loop uniform an external force of6.25×10−3N6.25×10-3Ntowards right must be applied.
∴W=(6.25×10−3N(0.5m)=3.125×10−3J

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