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

Options

A. Halved
B. Quadrupled
C. The same
D. Doubled

Explanation

To solve the problem, we need to analyze how the changes in the coil's parameters affect the power dissipated due to the induced current when the coil is placed in a time-varying magnetic field.

### Step-by-Step Solution:

1. **Understanding Induced EMF**:
The induced electromotive force (emf) in a coil placed in a time-varying magnetic field is given by Faraday's law of electromagnetic induction:
\[ \text{emf} (E) = -n \frac{d\Phi}{dt} \]
where \( n \) is the number of turns and \( \Phi \) is the magnetic flux.

2. **Magnetic Flux**:
The magnetic flux \( \Phi \) can be expressed as:
\[ \Phi = B \cdot A \]
where \( B \) is the magnetic field and \( A \) is the area of the coil. For a circular coil, the area \( A \) is given by:
\[ A = \pi r^2 \]

3. **Substituting for Area**:
Therefore, the induced emf can be rewritten as:
\[ E = -n \frac{d(B \cdot \pi r^2)}{dt} = -n \pi r^2 \frac{dB}{dt} \]

4. **Power Dissipated**:
The power \( P \) dissipated in the coil when it is short-circuited is given by:
\[ P = \frac{E^2}{R} \]
where \( R \) is the resistance of the coil.

5. **Resistance of the Coil**:
The resistance \( R \) of the coil can be expressed as:
\[ R = \rho \frac{L}{A_w} \]
where \( L \) is the length of the wire, \( A_w \) is the cross-sectional area of the wire, and \( \rho \) is the resistivity of the material. The length \( L \) of the wire is related to the number of turns \( n \) and the circumference of the coil:
\[ L = n \cdot 2\pi r \]
The cross-sectional area of the wire when the radius is doubled is:
\[ A_w = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \]
Thus, the new resistance \( R' \) when the radius is doubled becomes:
\[ R' = \rho \frac{n \cdot 2\pi r}{\frac{\pi d^2}{4}} = \frac{8\rho n r}{d^2} \]

6. **New Parameters**:
If the number of turns is halved, then \( n' = \frac{n}{2} \) and the radius is doubled \( r' = 2r \). The new area becomes:
\[ A' = \pi (2r)^2 = 4\pi r^2 \]
Thus, the new emf becomes:
\[ E' = -\frac{n}{2} \cdot 4\pi r^2 \frac{dB}{dt} = -2n \pi r^2 \frac{dB}{dt} \]

7. **New Power**:
The new power \( P' \) can be calculated as:
\[ P' = \frac{(E')^2}{R'} = \frac{(2n \pi r^2 \frac{dB}{dt})^2}{R'} \]
Substituting \( R' \):
\[ P' = \frac{(2n \pi r^2 \frac{dB}{dt})^2}{\frac{8\rho n r}{d^2}} = \frac{4n^2 \pi^2 r^4 \left(\frac{dB}{dt}\right)^2}{\frac{8\rho n r}{d^2}} = \frac{4n \pi^2 r^3 d^2 \left(\frac{dB}{dt}\right)^2}{8\rho} \]

8. **Comparing Powers**:
The original power \( P \) is:
\[ P = \frac{n^2 \pi^2 r^4 \left(\frac{dB}{dt}\right)^2}{R} \]
By comparing \( P' \) and \( P \), we can see that:
\[ P' = \frac{1}{4} P \]

### Final Answer:
Thus, the electrical power dissipated due to the current induced in the coil after halving the number of turns and doubling the radius of the wire is:
\[ P' = \frac{1}{4} P \]