Animated Solution for Physics - Electromagnetic Induction: The figure shows a square loop $L$ of side 5 cm which is connected to a network of resistances. The whole setup is moving towards right with a constant speed of $1{\text{ cm s}}^{-1}$. At some instant, a part of $L$ is in a uniform magnetic field of 1 T, perpendicular to the plane of the loop. If the resistance of $L$ is $1.7\Omega$, the current in the loop at that instant will be close to

Imagine a square conducting loop being pulled out of a magnetic field. As it moves, it acts like a battery, generating electricity that powers a complex network of resistors. This problem is a beautiful intersection of Faraday's Law of Electromagnetic Induction and Kirchhoff's Circuit Laws. Let's break down the physics and the math step-by-step.

The Engine of the Circuit

As the square loop $L$ moves to the right with a constant velocity $v$, its left arm cuts through the uniform magnetic field $B$. According to Faraday's Law, this motion induces an electromotive force (EMF).

Why only the left arm? The right arm is already outside the magnetic field region, so it experiences no flux change. The top and bottom arms are moving parallel to their own lengths; the magnetic force on the charges within them pushes perpendicular to the wire, not along it, so no EMF is generated along those segments.

The induced EMF $e$ is given by the motional EMF formula:

Substituting the given values: - Magnetic field, $B=1\text{ T}$ - Length of the arm, $l=5\text{ cm}=0.05\text{ m}$ - Velocity, $v=1\text{ cm/s}=0.01\text{ m/s}$

This $500\mu \text{V}$ acts as the driving voltage for our entire circuit.

The loop is connected to a diamond-shaped network of resistors. If you look closely, this is a classic Wheatstone Bridge. To solve it easily, we must first check if it is balanced.

Let's check the ratio of the resistances in the adjacent arms: - Left side ratio: $\frac{R_{AB}}{R_{BC}}=\frac{1\Omega }{2\Omega }=\frac{1}{2}$ - Right side ratio: $\frac{R_{AD}}{R_{DC}}=\frac{1\Omega }{2\Omega }=\frac{1}{2}$

Since the ratios are perfectly equal, the Wheatstone bridge is balanced. This is a crucial realization! In a balanced bridge, the nodes B and D are at the exact same electrical potential. Because there is no potential difference between them, absolutely zero current will flow through the central $3\Omega$ resistor. We can effectively remove it from our circuit analysis.

With the $3\Omega$ resistor gone, the bridge simplifies into two parallel branches: - The top branch has resistors $1\Omega$ and $2\Omega$ in series: $1+2=3\Omega$. - The bottom branch has resistors $1\Omega$ and $2\Omega$ in series: $1+2=3\Omega$.

The equivalent resistance of the bridge, $R_{\text{bridge}}$, is the parallel combination of these two $3\Omega$ branches:

Now, we must not forget the internal resistance of the loop itself, which is given as $1.7\Omega$. The total resistance of the entire circuit is:

Finally, we use Ohm's Law to find the current $I$ flowing through the loop:

Converting this to microamperes, we get $156.25\mu \text{A}$.

Looking at the options provided, $156.25\mu \text{A}$ is mathematically closest to option (c) $150\mu \text{A}$.

A Note on the Official Answer Key: The official answer key often lists (b) $170\mu \text{A}$ as the correct option. Why the discrepancy? This likely stems from a common mathematical error during the bridge simplification. If one mistakenly adds the left resistors ($1+1=2$) and the right resistors ($2+2=4$) and puts them in parallel, they would calculate $R_{\text{bridge}}=\frac{2\times 4}{2+4}\approx 1.33\Omega$. This incorrect resistance leads to a total resistance of $3.03\Omega$ and a current of $\approx 165\mu \text{A}$, which rounds to $170\mu \text{A}$. However, our rigorous step-by-step analysis confirms that $156.25\mu \text{A}$ is the true, mathematically sound result.