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 `2.00X 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. 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.
1.57 × 10 − 6 V
1.5 × 10 5 V
15.7 × 10 − 6 V
1.55 × 10 − 6 V

Options

A. `1.57xx10^(-6)V`
B. `1.5xx10^(5)V`
C. `15.7xx10^(-6)V`
D. `1.55xx10^(-6)V`

Explanation

To solve the problem, we need to find the average electromotive force (emf) induced in the iron part of the wheel as it rolls. Let's break down the steps:

### Step 1: Understand the Problem
We have a circular wheel with a radius of 10 cm, where the upper half is made of iron and the lower half is made of wood. A uniform magnetic field \( B \) of magnitude \( 2.00 \times 10^{-4} \, T \) exists above the central line. The wheel rolls for 2 seconds, during which the iron part comes down and the wooden part goes up.

### Step 2: Calculate the Area of the Iron Part
The area \( A \) of the iron part can be calculated as half of the area of the circle:
\[ A = \frac{1}{2} \pi r^2 \]
Substituting \( r = 0.1 \, m \) (10 cm):
\[ A = \frac{1}{2} \pi (0.1)^2 = \frac{1}{2} \pi (0.01) = \frac{\pi}{200} \, m^2 \]

### Step 3: Calculate the Magnetic Flux through the Iron Part
The magnetic flux \( \Phi \) through the iron part is given by:
\[ \Phi = B \cdot A \]
Substituting the values:
\[ \Phi = (2.00 \times 10^{-4} \, T) \cdot \left(\frac{\pi}{200} \, m^2\right) \]
Calculating this gives:
\[ \Phi = 2.00 \times 10^{-4} \cdot \frac{3.14}{200} = \frac{6.28 \times 10^{-4}}{200} = 3.14 \times 10^{-6} \, Wb \]

### Step 4: Calculate the Average Induced EMF
The average induced emf \( \mathcal{E} \) is given by the formula:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Since the flux changes from \( 0 \) to \( 3.14 \times 10^{-6} \, Wb \) in \( 2 \, s \):
\[ \mathcal{E} = -\frac{3.14 \times 10^{-6} \, Wb}{2 \, s} = -1.57 \times 10^{-6} \, V \]
The negative sign indicates the direction of the induced emf, but we are interested in the magnitude:
\[ \mathcal{E} = 1.57 \times 10^{-6} \, V \]

### Conclusion
The average emf induced during this period is \( 1.57 \times 10^{-6} \, V \).