0

Guest

Courses New

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 833 Points
3 years 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

Think You Can Provide A Better Answer ?

Other Related Questions on Magnetism

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More