Question

A metal disc of radius ‘R’ rotates with an angular velocity ‘ω’ about an axis perpendicular to its plane passing through its centre in a magnetic field of induction ‘B’ acting perpendicular to the plane of the disc. The induced e.m.f. between the rim and axis of the disc is (magnitude only)

• A. \(\frac {BωR}{2}\)
• B. \(\frac{Bω^2R^2}{2}\)
• C. \(\frac {BωR ^2}{2}\)
• D. \(\frac {Bω^2R}{2}\)

Answer 1

The Correct Option is C

Area A=πR²
Now differentiate the area
\(\frac {dA}{dt}\) = \(\frac {πR^2}{2π/ω}\)
\(\frac {dA}{dt}\)t = \(\frac {ωR^2}{2}\)
e.m.f = - B x \(\frac {dA}{dt}\)
e.m.f = \(\frac {BωR^2}{2}\)
So, the correct option is (C) \(\frac {BωR^2}{2}\)
 

Answer 2

To find the induced electromotive force (emf) between the rim and the axis of a rotating metal disc in a magnetic field, we can use Faraday's law of electromagnetic induction. The metal disc rotates in a uniform magnetic field, which creates a varying magnetic flux through the disc, inducing an emf.
Given:
- Radius of the disc, \( R \)
- Angular velocity of the disc, \( \omega \)
- Magnetic field induction, \( B \)
Steps to solve:
1. Consider an infinitesimal element:
 Consider an infinitesimal radial element of the disc at a distance \( r \) from the center with thickness \( dr \).
2. Linear velocity of the element:
The linear velocity \( v \) of the element at radius \( r \) is:
  \[ v = r \omega \]
3. Induced emf in the infinitesimal element:
 The emf \( d\mathcal{E} \) induced in the infinitesimal element due to its motion in the magnetic field is given by:
  \[ d\mathcal{E} = B \cdot v \cdot dr = B \cdot r \omega \cdot dr \]
4. Total induced emf:
 To find the total emf induced between the center and the rim of the disc, integrate \( d\mathcal{E} \) from \( r = 0 \) to \( r = R \):
  \[ \mathcal{E} = \int_{0}^{R} B \cdot r \omega \cdot dr \]
Performing the integration:
  \[ \mathcal{E} = B \omega \int_{0}^{R} r \cdot dr \]
  \[ \mathcal{E} = B \omega \left[ \frac{r^2}{2} \right]_{0}^{R} \]
  \[ \mathcal{E} = B \omega \cdot \frac{R^2}{2} \]
 Final Result:
The magnitude of the induced emf between the rim and the axis of the disc is Option (C):
\[ \boxed{\mathcal{E} = \frac{1}{2} B \omega R^2} \]