A Cu disc of radius 0.1 m rotates about its centre with 10 rps in a uniform magnetic field of 0.1 T. The emf induced across the radius of the disc is

Question

A Cu disc of radius 0.1 m rotates about its centre with 10 rps in a uniform magnetic field of 0.1 T. The emf induced across the radius of the disc is (A)  10π mV  (B)  20π mV  (C)  (π /10) V  (D)  (π /10) V

Tardigrade Admin · Accepted Answer

The induced emf between centre and rim of the rotating disc E=(1/2)Bω R2 =(1/2)× 0.1× 2π × 10× (0.1)2 =10π × 10-3V =10π mV

Q. A Cu disc of radius 0.1 m rotates about its centre with 10 rps in a uniform magnetic field of 0.1 T. The emf induced across the radius of the disc is

$10 π mV$

$20 π mV$

$\frac{π}{10} V$

$\frac{π}{10} V$

The induced emf between centre and rim of the rotating disc $E = \frac{1}{2} B ω R^{2}$ $= \frac{1}{2} \times 0.1 \times 2 π \times 10 \times (0.1)^{2}$ $= 10 π \times 10^{- 3} V$ $= 10 π mV$

Q. A Cu disc of radius 0.1 m rotates about its centre with 10 rps in a uniform magnetic field of 0.1 T. The emf induced across the radius of the disc is

10πmV

20πmV

10π​V

10π​V

The induced emf between centre and rim of the rotating disc E=21​BωR2 =21​×0.1×2π×10×(0.1)2 =10π×10−3V =10πmV

$10 π mV$

$20 π mV$

$\frac{π}{10} V$

$\frac{π}{10} V$

The induced emf between centre and rim of the rotating disc $E = \frac{1}{2} B ω R^{2}$ $= \frac{1}{2} \times 0.1 \times 2 π \times 10 \times (0.1)^{2}$ $= 10 π \times 10^{- 3} V$ $= 10 π mV$