Question

Figure shows a circular wheel of radius 10.0 cm whose upper half, shown dark in the figure, is made of iron and the lower half of wood. The two junctions are joined by an iron rod. A uniform magnetic field B of magnitude 2.00 × 10⁻⁴ T exists in the space above the central line as suggested by the figure. The wheel is set into pure rolling on the horizontal surface. If it takes 2.00 seconds for the iron part to come down and the wooden part to go up, find the average emf induced during this period.

Answer 1

Magnetic flux through the wheel (initially):-

\[\phi_1  = BA = \frac{2 \times {10}^{- 4} \times \pi \left( 0 . 1 \right)^2}{2}\]

\[         = \pi \times  {10}^{- 6}   Wb\]

As the wheel rotates, the wooden (non-metal) part of the wheel comes inside the magnetic field and the iron part of the wheel comes outside the magnetic field. Thus, the magnetic flux through the wheel becomes zero.

i.e. \[\phi_2 = 0\]

dt = 2 s

The average emf induced in the wheel is given by

\[e =  - \frac{d\phi}{dt}\]

\[       =  - \left( \frac{\phi_2 - \phi_1}{dt} \right)\]

\[       = \frac{\pi \times {10}^{- 6}}{2}\]

\[       = 1 . 57 \times  {10}^{- 6}   V\]