A current `I = 3.36(1 +2t) xx 10^(-2)` A increase at a steady state in a long staight wire. A small circular loop of radius `10^(-3)` m has its plane parallel to the wire and is placed at a distance of 1 m from the wire. The resistance of loop is `8.4 xx 10^(-4) (Omega)`. Find the approximate value of induced current in the loop.

Options

A. `5.024 xx 24^(-11)A`
B. `3.8 xx 24^(-11)A`
C. `2.75 xx 24^(-11)A`
D. `1.23 xx 24^(-11)A`

Explanation

To find the approximate value of the induced current in the loop, we can follow these steps:

### Step 1: Determine the rate of change of current (di/dt)
The current in the wire is given by:
\[ I = 3.36(1 + 2t) \times 10^{-2} \, \text{A} \]

To find the rate of change of current with respect to time, we differentiate \( I \) with respect to \( t \):
\[ \frac{dI}{dt} = \frac{d}{dt}[3.36(1 + 2t) \times 10^{-2}] = 3.36 \times 2 \times 10^{-2} = 6.72 \times 10^{-2} \, \text{A/s} \]

### Step 2: Calculate the magnetic field (B) at the location of the loop
The magnetic field \( B \) due to a long straight wire at a distance \( d \) is given by:
\[ B = \frac{\mu_0 I}{2 \pi d} \]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) and \( d = 1 \, \text{m} \).

Substituting \( I \) into the equation:
\[ B = \frac{4\pi \times 10^{-7} \times I}{2 \pi \times 1} = 2 \times 10^{-7} I \, \text{T} \]

### Step 3: Calculate the magnetic flux (Φ) through the loop
The magnetic flux \( \Phi \) through the loop is given by:
\[ \Phi = B \cdot A \]
where \( A \) is the area of the loop. The area \( A \) of a circular loop is:
\[ A = \pi r^2 \]
Given \( r = 10^{-3} \, \text{m} \):
\[ A = \pi (10^{-3})^2 = \pi \times 10^{-6} \, \text{m}^2 \]

Thus, the flux becomes:
\[ \Phi = B \cdot A = (2 \times 10^{-7} I) \cdot (\pi \times 10^{-6}) = 2\pi \times 10^{-13} I \, \text{Wb} \]

### Step 4: Find the induced EMF (ε)
The induced EMF \( \epsilon \) is given by Faraday's law of electromagnetic induction:
\[ \epsilon = -\frac{d\Phi}{dt} \]
Substituting the expression for \( \Phi \):
\[ \epsilon = -\frac{d}{dt}(2\pi \times 10^{-13} I) = -2\pi \times 10^{-13} \frac{dI}{dt} \]
Substituting \( \frac{dI}{dt} = 6.72 \times 10^{-2} \):
\[ \epsilon = -2\pi \times 10^{-13} \times 6.72 \times 10^{-2} \approx -4.5 \times 10^{-14} \, \text{V} \]

### Step 5: Calculate the induced current (I_induced)
Using Ohm's law, the induced current \( I_{\text{induced}} \) is given by:
\[ I_{\text{induced}} = \frac{\epsilon}{R} \]
where \( R = 8.4 \times 10^{-4} \, \Omega \):
\[ I_{\text{induced}} = \frac{-4.5 \times 10^{-14}}{8.4 \times 10^{-4}} \approx -5.36 \times 10^{-11} \, \text{A} \]

### Final Result
The approximate value of the induced current in the loop is:
\[ I_{\text{induced}} \approx 5.36 \times 10^{-11} \, \text{A} \]