A magnetic field induction is changing in magnitude at a constant rate `dB//dt` . A given mass `m` of copper is drawn into a wire of radius `alpha` and formed into a loop of radius `r` is placed perpendicular to the field. Show that induced current in the loop is given by `i=(m)/(4pipdelta)(dB)/(dt)`
`p` : resistivity, `delta` : density of copper.

Explanation

To solve the problem, we need to derive the expression for the induced current \( i \) in a loop formed from a given mass \( m \) of copper, which is subjected to a changing magnetic field \( B \) at a constant rate \( \frac{dB}{dt} \).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a mass \( m \) of copper that is drawn into a wire of radius \( \alpha \) and formed into a loop of radius \( r \).
- The loop is placed perpendicular to a magnetic field \( B \) that is changing at a rate \( \frac{dB}{dt} \).

2. **Calculate the Length of the Wire**:
- The length \( L \) of the wire used to form the loop is equal to the circumference of the loop:
\[ L = 2\pi r \]

3. **Calculate the Volume of Copper**:
- The volume \( V \) of the copper wire can be expressed in terms of its length and cross-sectional area:
\[ V = A \cdot L = \pi \alpha^2 \cdot L \]
- Substituting \( L \):
\[ V = \pi \alpha^2 (2\pi r) = 2\pi^2 r \alpha^2 \]

4. **Relate Mass, Density, and Volume**:
- The mass \( m \) of the copper can be expressed using its density \( \delta \):
\[ m = \delta V = \delta (2\pi^2 r \alpha^2) \]

5. **Calculate the Resistance of the Wire**:
- The resistance \( R \) of the wire can be calculated using the resistivity \( p \):
\[ R = \frac{pL}{A} = \frac{p(2\pi r)}{\pi \alpha^2} = \frac{2pr}{\alpha^2} \]

6. **Calculate the Induced EMF**:
- The induced EMF \( \mathcal{E} \) in the loop due to the changing magnetic field is given by Faraday's law:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
- The magnetic flux \( \Phi \) through the loop is:
\[ \Phi = B \cdot A = B \cdot \pi r^2 \]
- Therefore, the induced EMF becomes:
\[ \mathcal{E} = -\pi r^2 \frac{dB}{dt} \]

7. **Calculate the Induced Current**:
- The induced current \( i \) can be calculated using Ohm's law:
\[ i = \frac{\mathcal{E}}{R} \]
- Substituting the expressions for \( \mathcal{E} \) and \( R \):
\[ i = \frac{-\pi r^2 \frac{dB}{dt}}{\frac{2pr}{\alpha^2}} = -\frac{\pi r^2 \alpha^2}{2pr} \frac{dB}{dt} \]
- Simplifying this gives:
\[ i = -\frac{\pi r \alpha^2}{2p} \frac{dB}{dt} \]

8. **Substituting for \( r \)**:
- From the mass equation, we can express \( r \) in terms of \( m \):
\[ m = 2\pi^2 r \alpha^2 \delta \implies r = \frac{m}{2\pi^2 \alpha^2 \delta} \]
- Substitute \( r \) back into the current expression:
\[ i = -\frac{\pi \left(\frac{m}{2\pi^2 \alpha^2 \delta}\right) \alpha^2}{2p} \frac{dB}{dt} \]
- Simplifying this gives:
\[ i = -\frac{m}{4\pi p \delta} \frac{dB}{dt} \]

9. **Final Result**:
- Thus, the induced current \( i \) in the loop is given by:
\[ i = \frac{m}{4\pi p \delta} \frac{dB}{dt} \]