A magnetic field induction is changing in magnitude in a region at a constant rate `dB//dt`. A given mass `m` of copper drawn into a wire and formed into a loop is placed perpendicular to the field. If the values of specific resistance and density of copper are `rho` and `sigma` respectively, then the current in the loop is given by :

Options

A. (a) `(m)/(2pirhod)(dB)/(dt)`
B. (b) `(m)/(4pia^(2)r)(dB)/(dt)`
C. (c) `(m)/(4piad)(dB)/(dt)`
D. (d) `(m)/(4pirhod)(dB)/(dt)`

Explanation

To solve the problem, we need to determine the current in a loop made from a given mass of copper when placed in a changing magnetic field. Here’s the step-by-step solution:

### Step 1: Understand the given parameters
We have:
- A magnetic field induction changing at a rate of \( \frac{dB}{dt} \).
- A mass \( m \) of copper.
- The specific resistance (resistivity) of copper denoted as \( \rho \).
- The density of copper denoted as \( \sigma \).

### Step 2: Relate mass to volume
The mass of the copper wire can be expressed in terms of its volume and density:
\[ m = \sigma \cdot V \]
where \( V \) is the volume of the wire.

### Step 3: Express volume in terms of cross-sectional area and length
The volume \( V \) of the wire can also be expressed as:
\[ V = A \cdot L \]
where \( A \) is the cross-sectional area and \( L \) is the length of the wire.

### Step 4: Relate mass to length
Substituting the volume expression into the mass equation gives:
\[ m = \sigma \cdot A \cdot L \]
From this, we can express the length \( L \) of the wire as:
\[ L = \frac{m}{\sigma \cdot A} \]

### Step 5: Determine the induced EMF
The induced EMF \( E \) in the loop due to the changing magnetic field is given by:
\[ E = -\frac{d\Phi}{dt} \]
For a loop of area \( A \) in a magnetic field \( B \), the magnetic flux \( \Phi \) is:
\[ \Phi = B \cdot A \]
Thus,
\[ E = A \cdot \frac{dB}{dt} \]

### Step 6: Calculate the area of the loop
If we assume the loop is circular, the area \( A \) can be expressed in terms of the radius \( r \):
\[ A = \pi r^2 \]
Since the length of the wire \( L \) is the circumference of the loop:
\[ L = 2\pi r \implies r = \frac{L}{2\pi} \]
Substituting this into the area formula gives:
\[ A = \pi \left(\frac{L}{2\pi}\right)^2 = \frac{L^2}{4\pi} \]

### Step 7: Substitute area into EMF formula
Now substituting \( A \) back into the EMF equation:
\[ E = \frac{L^2}{4\pi} \cdot \frac{dB}{dt} \]

### Step 8: Calculate the resistance of the loop
The resistance \( R \) of the wire is given by:
\[ R = \frac{\rho \cdot L}{A} \]
Substituting for \( A \):
\[ R = \frac{\rho \cdot L}{\frac{L^2}{4\pi}} = \frac{4\pi \rho}{L} \]

### Step 9: Calculate the current using Ohm's Law
Using Ohm's Law, the current \( I \) is given by:
\[ I = \frac{E}{R} \]
Substituting the expressions for \( E \) and \( R \):
\[ I = \frac{\frac{L^2}{4\pi} \cdot \frac{dB}{dt}}{\frac{4\pi \rho}{L}} = \frac{L^3}{16\pi^2 \rho} \cdot \frac{dB}{dt} \]

### Step 10: Substitute \( L \) in terms of mass and density
From earlier, we have \( L = \frac{m}{\sigma A} \). Substitute this into the current equation:
\[ I = \frac{\left(\frac{m}{\sigma A}\right)^3}{16\pi^2 \rho} \cdot \frac{dB}{dt} \]

### Final Step: Simplify the expression
After simplification, we arrive at the final expression for the current \( I \):
\[ I = \frac{m}{4\pi \sigma \rho} \cdot \frac{dB}{dt} \]

### Conclusion
Thus, the current in the loop is given by:
\[ I = \frac{m}{4\pi \sigma \rho} \cdot \frac{dB}{dt} \]