Question

Figure shows a conducting disc rotating about its axis in a perpendicular magnetic field B. A resistor of resistance R is connected between the centre and the rim. Calculate the current in the resistor. Does it enter the disc or leave it at the centre? The radius of the disc is 5.0 cm, angular speed ω = 10 rad/s, B = 0.40 T and R = 10 Ω.

Answer 1

Given:-

Magnetic field perpendicular to the disc, B = 0.40 T

Angular speed, ω = 10 rad/s

Resistance, R = 10 Ω

Radius of the disc, r = 5 cm = 0.5 m

Let us consider a rod of length 0.05 m fixed at the centre of the disc and rotating with the same ω.

Now,

\[v = \frac{l}{2} \times \omega = \frac{0 . 05}{2} \times 10\]

\[e = Blv\]

\[       = 0 . 40 \times 0 . 05 \times \frac{0 . 05}{2} \times 10\]

\[       = 5 \times  {10}^{- 3}   V\]

\[i = \frac{e}{R}\]

\[     = \frac{5 \times {10}^{- 3}}{10} = 0 . 5  mA\]

As the disc is rotating in the anti-clockwise direction, the emf induced in the disc is such that the centre is at the higher potential and the periphery is at the lower potential. Thus, the current leaves from the centre.