To tackle this problem, we need to analyze the motion of the conducting rod and how it interacts with the magnetic field. The setup involves a conducting rod rotating in a magnetic field, which induces an electromotive force (emf) due to Faraday's law of electromagnetic induction. Let's break down the problem step by step.

Understanding the System

We have a conducting rod \( pq \) of mass \( m \) that rotates about a point \( O \) on a horizontal plane. The rod is connected to a resistance \( R \), and the entire system is placed in a uniform magnetic field \( \mathbf{B} \) that is perpendicular to the plane of the loop formed by the rod's rotation. The diameter of the circular path is \( l \), which gives us a radius \( r = \frac{l}{2} \).

Induced EMF and Current

As the rod rotates with an angular velocity \( \omega \), it cuts through the magnetic field lines, inducing an emf \( \mathcal{E} \). According to Faraday's law, the induced emf can be expressed as:

\(\mathcal{E} = B \cdot l \cdot v\)

Here, \( v \) is the linear velocity of the rod, which can be calculated as:

\(v = r \cdot \omega\)

Substituting \( v \) into the equation for emf gives:

\(\mathcal{E} = B \cdot l \cdot (r \cdot \omega)\)

Since \( r = \frac{l}{2} \), we can simplify this to:

\(\mathcal{E} = \frac{B \cdot l^2 \cdot \omega}{2}\)

Current as a Function of Time

The current \( I(t) \) flowing through the resistance \( R \) can be calculated using Ohm's law:

\(I(t) = \frac{\mathcal{E}}{R}\)

Substituting the expression for emf, we find:

\(I(t) = \frac{B \cdot l^2 \cdot \omega}{2R}\)

Since the angular velocity \( \omega \) is constant, the current remains constant over time, so we can denote it as:

\(I(t) = I_0 = \frac{B \cdot l^2 \cdot \omega}{2R}\)

Total Charge Flowing Through the Resistance

The total charge \( Q \) that flows through the resistance over time \( t \) can be calculated using the relationship:

\(Q = I \cdot t\)

Substituting the expression for current, we have:

\(Q(t) = I_0 \cdot t = \frac{B \cdot l^2 \cdot \omega}{2R} \cdot t\)

Heat Generated in the Circuit at Infinite Time

The heat generated \( H \) in the circuit can be calculated using Joule's law, which states that the heat produced in a resistor is given by:

\(H = I^2 \cdot R \cdot t\)

Substituting for \( I \), we find:

\(H = \left(\frac{B \cdot l^2 \cdot \omega}{2R}\right)^2 \cdot R \cdot t\)

As \( t \) approaches infinity, the heat generated can be expressed as:

\(H = \frac{B^2 \cdot l^4 \cdot \omega^2}{4R} \cdot t\)

Summary of Results

• Magnitude of Current: \( I(t) = \frac{B \cdot l^2 \cdot \omega}{2R} \)
• Total Charge Flow: \( Q(t) = \frac{B \cdot l^2 \cdot \omega}{2R} \cdot t \)
• Heat Generated: \( H = \frac{B^2 \cdot l^4 \cdot \omega^2}{4R} \cdot t \) as \( t \to \infty \)

This analysis provides a comprehensive understanding of the behavior of the conducting rod in the magnetic field, the current induced, the charge flow, and the heat generated over time. If you have any further questions or need clarification on any part, feel free to ask!