A close loop 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 keeping the radius of the loop unchanged, the electrical power dissipated would be :

Options

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

Answer

To solve the problem step by step, we need to analyze how the changes in the number of turns and the wire radius affect the electrical power dissipated in the loop placed in a time-varying magnetic field.

### Step 1: Understand the Power Dissipation Formula
The electrical power \( P \) dissipated in the coil can be expressed as:
\[ P = \frac{E^2}{R} \]
where \( E \) is the induced EMF and \( R \) is the resistance of the coil.

### Step 2: Calculate the Induced EMF
The induced EMF \( E \) in the coil due to a changing magnetic field is given by:
\[ E = -\frac{d\Phi}{dt} \]
where \( \Phi \) is the magnetic flux through the coil. For a coil with \( n \) turns, the magnetic flux is:
\[ \Phi = n \cdot B \cdot A \]
where \( A \) is the area of the loop and \( B \) is the magnetic field.

### Step 3: Area of the Loop
The area \( A \) of the loop is given by:
\[ A = \pi R^2 \]
where \( R \) is the radius of the loop.

### Step 4: Resistance of the Wire
The resistance \( R \) of the wire can be calculated using:
\[ R = \frac{\rho L}{A_w} \]
where \( \rho \) is the resistivity of the wire material, \( L \) is the length of the wire, and \( A_w \) is the cross-sectional area of the wire. The cross-sectional area of the wire with radius \( r \) is:
\[ A_w = \pi r^2 \]

### Step 5: Changes in Parameters
1. The number of turns \( n \) is quadrupled: \( n' = 4n \).
2. The radius of the wire is halved: \( r' = \frac{r}{2} \).

### Step 6: New Resistance Calculation
The new length of the wire \( L' \) when the number of turns is quadrupled is:
\[ L' = 4L \]
The new cross-sectional area of the wire becomes:
\[ A_w' = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4} \]
Thus, the new resistance \( R' \) is:
\[ R' = \frac{\rho L'}{A_w'} = \frac{\rho (4L)}{\frac{\pi r^2}{4}} = \frac{16\rho L}{\pi r^2} = 16R \]

### Step 7: New Induced EMF Calculation
The new induced EMF \( E' \) becomes:
\[ E' = -\frac{d\Phi'}{dt} = -\frac{d(n' \cdot B \cdot A)}{dt} = -\frac{d(4n \cdot B \cdot A)}{dt} = 4E \]

### Step 8: New Power Calculation
Now substituting the new values into the power formula:
\[ P' = \frac{(E')^2}{R'} = \frac{(4E)^2}{16R} = \frac{16E^2}{16R} = \frac{E^2}{R} = P \]

### Conclusion
The new electrical power dissipated \( P' \) remains the same as the original power \( P \).

### Final Answer
The electrical power dissipated would remain the same.

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