NEET + JEEAdvanced
A conducting rod of length 0.50 m rotates in a horizontal plane about…
Question
A conducting rod of length 0.50 m rotates in a horizontal plane about a vertical axis through one of its ends with constant angular speed 20 rad/s. A uniform magnetic field of 0.40 T is directed perpendicular to the plane of rotation. Find the EMF (in volts) induced between the two ends of the rod.
✓ Verified answer: 1checked by our engine — not a guess
Step-by-step solution
For a rod rotating about one end, the motional EMF is (1/2) B omega L^2. EMF = 0.5 x 0.40 x 20 x (0.50)^2 = 0.5 x 0.40 x 20 x 0.25 = 1.0 V.
Final answer1.0
Stuck on a problem like this?
Paste any JEE or NEET question — verified working, a confidence %, and an honest “not sure” instead of a bluff.
Solve my doubt →More Neet Magnetism Emi solutions
A cyclotron accelerates protons (mass 1.67 x 10^-27 kg, charge 1.6 x…NEET + JEE · PhysicsA charged particle with charge-to-mass ratio q/m = 1.0 x 10^8 C/kg…NEET + JEE · PhysicsTwo long straight parallel wires carry currents 10 A and 15 A in…NEET + JEE · PhysicsA square coil of side 0.10 m has 50 turns and carries a current of…NEET + JEE · PhysicsA long solenoid has 1000 turns per metre. A short secondary coil of…NEET + JEE · PhysicsA series LCR circuit has resistance 30 ohm, inductive reactance 50…NEET + JEE · PhysicsA charged particle with charge-to-mass ratio q/m = 5.0 x 10^7 C/kg…NEET + JEE · PhysicsA conducting rod of length 0.40 m slides without friction on two…NEET + JEE · Physics