A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would be

Options

A. halved
B. the same
C. doubled
D. quadrupled

Explanation

To solve the problem, we need to analyze how the electrical power dissipated in a short-circuited coil changes when the number of turns is quadrupled and the wire radius is halved.

### Step-by-Step Solution:

1. **Understanding Power Dissipation**:
The electrical power \( P \) dissipated in a coil due to the induced current can be expressed as:
\[ P = \frac{E^2}{R} \]
where \( E \) is the induced electromotive force (emf) and \( R \) is the resistance of the coil.

2. **Finding the Induced EMF**:
According to Faraday's law of electromagnetic induction, the induced emf \( E \) in the coil is given by:
\[ E = -\frac{d\Phi}{dt} \]
where \( \Phi \) is the magnetic flux. For a coil with \( n \) turns and area \( A \), the induced emf can be expressed as:
\[ E \propto n \cdot A \cdot \frac{dB}{dt} \]

3. **Calculating the Area**:
The area \( A \) of the coil is given by:
\[ A = \pi r^2 \]
where \( r \) is the radius of the coil.

4. **New Configuration**:
- The number of turns is quadrupled: \( n' = 4n \)
- The radius is halved: \( r' = \frac{r}{2} \)

5. **Calculating New Area**:
The new area \( A' \) becomes:
\[ A' = \pi (r')^2 = \pi \left(\frac{r}{2}\right)^2 = \pi \frac{r^2}{4} = \frac{A}{4} \]

6. **Finding New Induced EMF**:
The new induced emf \( E' \) can be expressed as:
\[ E' \propto n' \cdot A' \cdot \frac{dB}{dt} = (4n) \cdot \left(\frac{A}{4}\right) \cdot \frac{dB}{dt} = n \cdot A \cdot \frac{dB}{dt} = E \]
Thus, \( E' = E \).

7. **Calculating Resistance**:
The resistance \( R \) of the coil is given by:
\[ R = \rho \frac{L}{A_w} \]
where \( L \) is the length of the wire and \( A_w \) is the cross-sectional area of the wire. The cross-sectional area of the wire when the radius is halved is:
\[ A_w' = \pi \left(\frac{r_w}{2}\right)^2 = \frac{A_w}{4} \]
The length of the wire increases with the number of turns, so:
\[ L' = 4L \]
Therefore, the new resistance becomes:
\[ R' = \rho \frac{4L}{\frac{A_w}{4}} = 16 \cdot \rho \frac{L}{A_w} = 16R \]

8. **Calculating New Power**:
The new power \( P' \) is given by:
\[ P' = \frac{E'^2}{R'} = \frac{E^2}{16R} \]

9. **Finding the Ratio of Powers**:
The ratio of the new power to the original power is:
\[ \frac{P'}{P} = \frac{\frac{E^2}{16R}}{\frac{E^2}{R}} = \frac{1}{16} \]

### Conclusion:
The electrical power dissipated in the coil when the number of turns is quadrupled and the wire radius is halved will be \( \frac{1}{16} \) of the original power.