Question

A horizontal conducting loop of radius $R$ is fixed in air. A uniformly charged rod having charge $Q$ on it is held vertically above the conducting loop at a height $h$ above it. The rod is released and it begins to fall along the axis of the loop. Calculate the emf induced in the conducting loop.

Answer 1

Zero

Answer 2

To calculate the induced electromotive force (EMF) in the conducting loop, we need to use Faraday's Law of Electromagnetic Induction, which states:

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

where $\Phi_B$ is the magnetic flux through the loop. The magnetic flux is defined as:

$$\Phi_B = \int \vec{B} \cdot d\vec{A}$$

Here, $\vec{B}$ is the magnetic field produced by the moving charged rod, and $d\vec{A}$ is the infinitesimal area vector of the conducting loop.

1. Magnetic field due to the moving charged rod:

   The uniformly charged rod, moving with velocity $v$ along its length, can be considered as a current-carrying wire. If the linear charge density of the rod is $\lambda = Q/L$ (where $L$ is the length of the rod), then the equivalent current is $I = \lambda v$.

   The rod falls along the axis of the conducting loop. Let's assume the axis of the loop (and the path of the rod) is the z-axis, and the loop lies in the xy-plane.

   The magnetic field lines produced by a straight current-carrying wire (the falling rod) are concentric circles around the wire. Therefore, at any point in space, the magnetic field vector $\vec{B}$ due to the rod is tangential to these circles.

2. Orientation of the magnetic field relative to the loop's area vector:

   The conducting loop is in the xy-plane, centered on the z-axis. Its area vector $d\vec{A}$ is perpendicular to the plane of the loop, meaning $d\vec{A}$ points along the z-axis (e.g., $d\vec{A} = dA \hat{k}$).

   As established, the magnetic field $\vec{B}$ produced by the current along the z-axis is circular, lying in planes perpendicular to the z-axis (i.e., in the xy-plane).

   Therefore, the magnetic field vector $\vec{B}$ is always perpendicular to the area vector $d\vec{A}$ of the loop at every point on the loop.

3. Calculation of magnetic flux:

   Since $\vec{B}$ is perpendicular to $d\vec{A}$ everywhere on the loop, their dot product $\vec{B} \cdot d\vec{A}$ is always zero.

   $$\vec{B} \cdot d\vec{A} = B dA \cos(90^\circ) = 0$$

   Consequently, the total magnetic flux through the loop is:

   $$\Phi_B = \int \vec{B} \cdot d\vec{A} = \int 0 \, dA = 0$$

4. Calculation of induced EMF:

   Since the magnetic flux $\Phi_B$ through the loop is zero at all times, its time derivative is also zero:

   $$\frac{d\Phi_B}{dt} = \frac{d(0)}{dt} = 0$$

   According to Faraday's Law, the induced EMF is:

   $$\mathcal{E} = -\frac{d\Phi_B}{dt} = 0$$

Therefore, no EMF is induced in the conducting loop. This is because, regardless of the rod's position or velocity (even if it's accelerating due to gravity, causing the current and magnetic field strength to change), the magnetic field lines are always perpendicular to the area vector of the loop, resulting in zero magnetic flux.

Explanation of the solution:

The magnetic field produced by the falling charged rod (which acts as a current) forms concentric circles around the axis of the loop. The area vector of the loop is parallel to this axis. Since the magnetic field lines are perpendicular to the axis, they are perpendicular to the loop's area vector. Thus, the magnetic flux through the loop ($\Phi_B = \int \vec{B} \cdot d\vec{A}$) is always zero. According to Faraday's law ($\mathcal{E} = -d\Phi_B/dt$), if the flux is always zero, its rate of change is also zero, leading to zero induced EMF.