A uniform magnetic field B exists in a direction perpendicular to the plane of a square loop made of

Q: Why do we divide the length by 4 to get the side of the square?

The total length of the wire ($30 \text{ cm}$) is bent to form the perimeter of the square. Since a square has 4 equal sides, the perimeter is $4a = L$. Therefore, each side $a$ is exactly $L/4$.

Q: I used the area of the square loop to calculate the resistance. Why is my answer wrong?

This is the most common trap! The resistance of a wire depends on how thick the wire itself is, not how much space the loop encloses. You must use the microscopic cross-sectional area of the wire ($\pi r^2$) for resistance, and the macroscopic area of the square ($a^2$) for the magnetic flux.

Q: Does the initial direction of the magnetic field matter for finding the magnitude of the current?

No, the magnitude of the induced EMF (and thus the current) depends strictly on the *rate of change* of the magnetic field ($dB/dt$), not its initial direction or magnitude. The direction of the field would only matter if you were asked to find the direction of the induced current using Lenz's Law.

Q: What would happen if the magnetic field was very strong but constant?

If the magnetic field is constant, then its rate of change $dB/dt$ is exactly zero. According to Faraday's Law, no matter how strong the field is, if it isn't changing, there is no induced EMF and therefore zero induced current.

Q: Why do we need to convert millimeters and centimeters to meters?

Physics formulas rely on consistent SI units. The resistivity is given in $\Omega\text{m}$ and the magnetic field rate is in Tesla per second (which is derived from standard SI units). If you leave lengths in cm or mm, your final current will be off by several orders of magnitude.

Question

A uniform magnetic field $B$ exists in a direction perpendicular to the plane of a square loop made of a metal wire. The wire has a diameter of $4 	ext{ mm}$ and a total length of $30 	ext{ cm}$. The magnetic field changes with time at a steady rate $dB/dt = 0.032 	ext{ Ts}^{-1}$. The induced current in the loop is close to (Take, resistivity of the metal wire $= 1.23 	imes 10^{-8} \Omega	ext{m}$)

Options:

(A) 0.61 A

(B) 0.43 A

(C) 0.53 A

(D) 0.34 A

Sigma Percentile · Accepted Answer

### The Tale of Two Areas: Unraveling Induced Current

Imagine you are handed a piece of metal wire, exactly $30 	ext{ cm}$ long, and asked to bend it into a perfect square. You place this square on a table, and suddenly, an invisible magnetic field starts piercing through it, growing stronger every second. What happens next is pure magic—or rather, pure physics. A current begins to flow through the wire out of nowhere!

This problem is a beautiful interplay between the macroscopic geometry of the loop and the microscopic properties of the wire itself. Let's break it down step by step.

### The Macroscopic World: Faraday's Law

First, we need to understand how much "push" or Electromotive Force (EMF) is being generated. According to Faraday's Law of Induction, the induced EMF is equal to the rate of change of magnetic flux.

Mathematically, $E = \left| \frac{d\Phi}{dt} ight|$. Since the magnetic field is perpendicular to the loop, the flux is simply $\Phi = B \cdot A_{	ext{loop}}$. Because the area of the loop isn't changing, we can pull it out of the derivative:

$$E = A_{	ext{loop}} \frac{dB}{dt}$$

But what is the area of our loop? The total length of the wire is $L = 0.3 	ext{ m}$. Since it's bent into a square, each side is $a = \frac{L}{4} = 0.075 	ext{ m}$. Therefore, the macroscopic area capturing the magnetic flux is $A_{	ext{loop}} = (0.075)^2 	ext{ m}^2$.

Given the rate of change of the magnetic field $\frac{dB}{dt} = 0.032 	ext{ T/s}$, the induced EMF is:

$$E = (0.075)^2 	imes 0.032 = 1.8 	imes 10^{-4} 	ext{ V}$$

### The Microscopic World: Resistance of the Wire

Now that we have the EMF, we need to know how much the wire resists the flow of electrons. This is where many students fall into a classic trap! They reuse the area of the square loop. **Do not do this.** The resistance depends on the *cross-sectional area* of the wire itself, not the area enclosed by the loop.

The formula for resistance is $R = ho \frac{L}{A_{	ext{wire}}}$.

The wire has a diameter of $4 	ext{ mm}$, meaning its radius is $r = 2 	imes 10^{-3} 	ext{ m}$. The cross-sectional area is:

$$A_{	ext{wire}} = \pi r^2 = \pi (2 	imes 10^{-3})^2 \approx 12.56 	imes 10^{-6} 	ext{ m}^2$$

Plugging in the given resistivity $ho = 1.23 	imes 10^{-8} \Omega	ext{m}$ and the total length $L = 0.3 	ext{ m}$:

$$R = \frac{1.23 	imes 10^{-8} 	imes 0.3}{12.56 	imes 10^{-6}} \approx 2.94 	imes 10^{-4} \Omega$$

### Bringing it Together: Ohm's Law

Finally, we unite the macroscopic EMF and the microscopic resistance using Ohm's Law to find the induced current:

$$I = \frac{E}{R} = \frac{1.8 	imes 10^{-4}}{2.94 	imes 10^{-4}} \approx 0.61 	ext{ A}$$

The induced current is approximately **$0.61 	ext{ A}$**, which perfectly matches option (a).

**A thought experiment for you:** What if we took the exact same $30 	ext{ cm}$ wire and bent it into a circle instead of a square? For a given perimeter, a circle encloses the maximum possible area. A larger area means more magnetic flux is captured, leading to a higher induced EMF and, consequently, a higher induced current! Physics is all about recognizing these elegant geometric optimizations.

Select Answer:

Visualized Solution

Geometry of the Loop and Wire

Faraday's Law of Induction

Calculating Induced EMF

Resistance of the Wire

Calculating Resistance

Ohm's Law for Induced Current

Conclusion & The Way Forward

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

The Tale of Two Areas

The Macroscopic World

The Microscopic World

Bringing it Together

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