Question

A conducting loop (as shown) has total resistance R. A uniform magnetic field $B = \gamma t$ is applied perpendicular outwards to plane of the loop where $\gamma$ is a constant and $t$ is time as shown. The induced current flowing through loop is

Answer 1

$\frac{\gamma (a^2 + b^2)}{R}$

Answer 2

The problem asks for the induced current in a conducting loop due to a time-varying magnetic field.

1. Calculate the total area of the loop:
   The loop consists of two square sections.
   Area of the first square (top) = $b \times b = b^2$.
   Area of the second square (bottom) = $a \times a = a^2$.
   The total area ($A_{total}$) through which the magnetic field passes is the sum of these areas:
   $A_{total} = a^2 + b^2$

2. Calculate the magnetic flux ($\Phi_B$):
   The magnetic field $B = \gamma t$ is uniform and applied perpendicular outwards to the plane of the loop.
   The magnetic flux is given by $\Phi_B = B \cdot A_{total}$.
   Substituting the given values:
   $\Phi_B = (\gamma t)(a^2 + b^2)$

3. Calculate the induced electromotive force (EMF) using Faraday's Law:
   Faraday's Law states that the induced EMF ($\mathcal{E}$) is the negative rate of change of magnetic flux:
   $\mathcal{E} = - \frac{d\Phi_B}{dt}$
   $\mathcal{E} = - \frac{d}{dt} [\gamma t (a^2 + b^2)]$
   Since $\gamma$, $a$, and $b$ are constants:
   $\mathcal{E} = - \gamma (a^2 + b^2) \frac{dt}{dt}$
   $\mathcal{E} = - \gamma (a^2 + b^2)$
   The magnitude of the induced EMF is:
   $|\mathcal{E}| = \gamma (a^2 + b^2)$

4. Calculate the induced current (I) using Ohm's Law:
   The total resistance of the loop is given as $R$.
   According to Ohm's Law, the induced current is:
   $I = \frac{|\mathcal{E}|}{R}$
   Substituting the magnitude of EMF:
   $I = \frac{\gamma (a^2 + b^2)}{R}$

The direction of the current can be found using Lenz's Law. Since the outward magnetic field is increasing, the induced current will flow in a clockwise direction to produce an inward magnetic field, opposing the change. The question only asks for the magnitude of the current.

The final answer is $\frac{\gamma (a^2 + b^2)}{R}$

Explanation of the solution:
The total area of the loop is $A = a^2 + b^2$. The magnetic flux through the loop is $\Phi_B = B \cdot A = \gamma t (a^2 + b^2)$. By Faraday's Law, the induced EMF is $\mathcal{E} = -\frac{d\Phi_B}{dt} = -\gamma (a^2 + b^2)$. The magnitude of the induced EMF is $|\mathcal{E}| = \gamma (a^2 + b^2)$. Using Ohm's Law, the induced current is $I = \frac{|\mathcal{E}|}{R} = \frac{\gamma (a^2 + b^2)}{R}$.

Answer:
The induced current flowing through the loop is $\frac{\gamma (a^2 + b^2)}{R}$.