Induced Current Calculation in Square Frame

This problem requires us to calculate the induced current in a square frame of copper wire when subjected to a time-varying magnetic field. We will use principles like Faraday's Law and Ohm's Law.

Magnetic Field and Frame Parameters

Here's a summary of the given information:

• Frame Shape: Square
• Material: Copper wire
• Wire diameter ($d$): $2 \ mm = 2 \times 10^{-3} \ m$
• Total wire length ($L_{total}$): $40 \ cm = 0.4 \ m$
• Rate of magnetic field change ($\frac{dB}{dt}$): $0.02 \ Ts^{-1}$
• Copper resistivity ($\rho$): $1.7 \times 10^{-8} \ \Omega m$
• Magnetic field orientation: Perpendicular to the frame's plane

Solution Steps for Induced Current

1. Calculating the Frame Area

The total length of the wire ($L_{total}$) represents the perimeter of the square frame. Let the side length of the square be '$a$'.

Perimeter = $4a = L_{total}$

Given $L_{total} = 0.4 \ m$:

$4a = 0.4 \ m \implies a = \frac{0.4 \ m}{4} = 0.1 \ m$

The area ($A$) of the square frame is calculated as:

$A = a^2 = (0.1 \ m)^2 = 0.01 \ m^2$

2. EMF Calculation via Faraday's Law

Faraday's Law of Induction states that the induced EMF ($\mathcal{E}$) is equal to the negative rate of change of magnetic flux ($\Phi_B$).

Magnetic Flux, $\Phi_B = B \cdot A$, since the field is perpendicular to the area.

Therefore, the induced EMF is:

$\mathcal{E} = -\frac{d\Phi_B}{dt} = -A \frac{dB}{dt}$

We consider the magnitude of the EMF:

$|\mathcal{E}| = A \frac{dB}{dt}$

Plugging in the values:

$|\mathcal{E}| = (0.01 \ m^2) \times (0.02 \ Ts^{-1}) = 0.0002 \ V = 2 \times 10^{-4} \ V$

3. Resistance Calculation for Copper Wire

The resistance ($R$) of the copper wire is determined using $R = \rho \frac{L}{A_{wire}}$.

Wire length $L = 0.4 \ m$.

Wire radius $r = \frac{d}{2} = \frac{2 \times 10^{-3} \ m}{2} = 1 \times 10^{-3} \ m$.

Cross-sectional area of the wire, $A_{wire} = \pi r^2$:

$A_{wire} = \pi (1 \times 10^{-3} \ m)^2 = \pi \times 10^{-6} \ m^2$

Calculating resistance:

$R = (1.7 \times 10^{-8} \ \Omega m) \times \frac{0.4 \ m}{\pi \times 10^{-6} \ m^2}$

$R = \frac{1.7 \times 0.4}{\pi} \times 10^{-2} \ \Omega = \frac{0.68}{\pi} \times 10^{-2} \ \Omega$

Using $\pi \approx 3.14159$:

$R \approx \frac{0.68}{3.14159} \times 10^{-2} \ \Omega \approx 0.2166 \times 10^{-2} \ \Omega \approx 2.166 \times 10^{-3} \ \Omega$

4. Current Calculation using Ohm's Law

According to Ohm's Law, the induced current ($I_{induced}$) is the induced EMF divided by the total resistance.

$I_{induced} = \frac{|\mathcal{E}|}{R}$

Substituting the values:

$I_{induced} = \frac{2 \times 10^{-4} \ V}{2.166 \times 10^{-3} \ \Omega}$

$I_{induced} \approx 0.923 \times 10^{-1} \ A \approx 9.23 \times 10^{-2} \ A$

Final Induced Current Value

The calculated value for the induced current is approximately $9.23 \times 10^{-2} \ A$. This value is closest to option 2, which is $9.3 \times 10^{-2} \ A$.