Electro Magnetic Induction And Alternating Currents Question 518

Question: A rod OA of length l is rotating (about end 0) over a conducting ring in crossed magnetic field B with constant angular velocity co as shown in figure

Options:

A) Current flowing through the rod is Bω,l2R

B) Magnetic force acting on the rod is 3B2ω,l34R

C) Torque due to magnetic force acting on the rod is 3B2ω,l48R

D) Magnitude of external force that acts perpendicularly at the end of the rod to maintain the constant angular speed is 3B2ω,l35R .

Show Answer

Answer:

Correct Answer: C

Solution:

  • I=ε2R3=3ε2R

    =32R×12Bωl2 =3Bωl24R

    Magnetic force F=3Bωl24R×l×B=3B2ωl34R

    τ=3B2ωl34R×l2=3B2ωl48R

    Force to be applied at the end =3B2ωl38R .



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