Question

A circular loop of radius r of net resistance R is placed perpendicular to a uniform magnetic field B. The radius of the loop decreases with rate $(\alpha)$ m/s. The induced current in the loop is $\frac{\pi r \alpha B}{R}$ ampere. Find x

Answer 1

2

Answer 2

To determine the value of 'x', we first need to derive the expression for the induced current in the circular loop using Faraday's Law of Electromagnetic Induction.

1. Magnetic Flux (Φ): The magnetic field B is uniform and perpendicular to the plane of the circular loop. The area of the loop is $A = \pi r^2$. The magnetic flux ($\Phi$) through the loop is given by: $\Phi = \vec{B} \cdot \vec{A} = B A \cos\theta$ Since the magnetic field is perpendicular to the plane of the loop, the angle $\theta$ between the magnetic field vector ($\vec{B}$) and the area vector ($\vec{A}$) is $0^\circ$ (or $180^\circ$). Assuming $0^\circ$, $\cos 0^\circ = 1$. $\Phi = B (\pi r^2)$

2. Induced Electromotive Force (EMF, ε): According to Faraday's Law of Electromagnetic Induction, the induced EMF is given by the negative rate of change of magnetic flux: $\varepsilon = -\frac{d\Phi}{dt}$ Substitute the expression for $\Phi$: $\varepsilon = -\frac{d}{dt}(B \pi r^2)$ Since B and $\pi$ are constants, we can take them out of the derivative: $\varepsilon = -B \pi \frac{d}{dt}(r^2)$ Using the chain rule, $\frac{d}{dt}(r^2) = 2r \frac{dr}{dt}$. $\varepsilon = -B \pi (2r \frac{dr}{dt})$ The problem states that the radius of the loop decreases with a rate $(\alpha)$ m/s. Therefore, $\frac{dr}{dt} = -\alpha$ (the negative sign indicates that the radius is decreasing). Substitute this into the EMF equation: $\varepsilon = -2 \pi B r (-\alpha)$ $\varepsilon = 2 \pi B r \alpha$ The magnitude of the induced EMF is $|\varepsilon| = 2 \pi B r \alpha$.

3. Induced Current (I): According to Ohm's Law, the induced current (I) in the loop is the induced EMF divided by the net resistance (R) of the loop: $I = \frac{|\varepsilon|}{R}$ $I = \frac{2 \pi B r \alpha}{R}$

4. Comparing with the given expression: The problem states that "The induced current in the loop is $\frac{\pi r \alpha B}{R}$ ampere. Find x". This phrasing implies that the given expression for the current might be missing a numerical factor 'x'. Let's assume the question implicitly asks for 'x' such that the correct induced current is $x$ times the given expression, or the given expression is $x$ times the correct one. The most common interpretation for such questions is that the correct formula is $x$ times the given formula or the derived formula is $x$ times the given formula. Our derived current is $I_{derived} = \frac{2 \pi r \alpha B}{R}$. The current mentioned in the question is $I_{given} = \frac{\pi r \alpha B}{R}$. By comparing the two expressions: $I_{derived} = 2 \times \left( \frac{\pi r \alpha B}{R} \right)$ So, $I_{derived} = 2 \times I_{given}$. If the question is asking for the factor 'x' by which the given expression needs to be multiplied to get the correct induced current, then $x=2$.

The final answer is $\boxed{2}$.

Explanation of the solution: The magnetic flux through the loop is $\Phi = B(\pi r^2)$. By Faraday's law, the induced EMF is $\varepsilon = -\frac{d\Phi}{dt} = -B\pi \frac{d(r^2)}{dt} = -B\pi (2r \frac{dr}{dt})$. Given that the radius decreases at a rate $\alpha$, $\frac{dr}{dt} = -\alpha$. Substituting this, $\varepsilon = -2\pi Br(-\alpha) = 2\pi Br\alpha$. The induced current is $I = \frac{|\varepsilon|}{R} = \frac{2\pi Br\alpha}{R}$. Comparing this derived current with the given expression $\frac{\pi r \alpha B}{R}$, we find that the derived current is twice the given expression. Therefore, the value of x is 2.