Questions
A long conducting wire AH is moved over a conducting triangular wire CDE with a constant velocity v in a uniform magnetic field directed into the plane of the paper. Resistance per unit length of each wire is . Then
detailed solution
Correct option is D
As the wire AH slides, the enclosed area increases, so the magnetic flux in ⊗ direction through the coil is increasing. Therefore, induced current will produce magnetic field in ⊙ direction. Thus, the current in the loop is anticlockwise. Magnitude of induced current at any instant of time isi=eR=Bv(FG)ρ(FG+GD+DF)When the wire AH moves downwards FG, GD and DF all increase in the same ratio. Therefore, i is constant.Talk to our academic expert!
Similar Questions
A square wire loop, of side a has a long straight wire carrying a constant current i that passes through the centre of the loop and perpendicular to its plane as shown in Fig.
The current induced in the square loop is
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
