Two concentric circular coils C1 and C2 are placed in the xy-plane. C1 has 500 turns and radius of 1

Q: Why do we assume the magnetic field is uniform over the inner coil?

Because $r_1 \ll r_2$ (1 cm vs 20 cm), the variation of the magnetic field near the center of the large coil is negligible over the small area of the inner coil.

Q: Why did we use $\pi^2 \approx 10$?

In many JEE physics problems, taking $\pi^2 \approx 10$ simplifies the calculation significantly and leads to the intended integer answer. Always look out for this if you are stuck with a $\pi^2$ in a numerical type question.

Q: What if the inner coil was larger?

If the inner coil's radius was comparable to the outer coil, we would have to integrate the magnetic field over the area of the inner coil, which involves complex elliptic integrals and is beyond the JEE syllabus.

Q: Does the resistance of the coils matter here?

No, because we are only asked for the induced electromotive force (emf), not the induced current. Resistance would only be needed if we had to find the current.

Q: Can we use the concept of mutual inductance here?

Absolutely! You can first find the mutual inductance $M = \frac{\mu_0 N_1 N_2 \pi r_1^2}{2 r_2}$ and then use $e = M \frac{dI}{dt}$. It's exactly the same mathematical process.

Question

Two concentric circular coils $C_1$ and $C_2$ are placed in the xy-plane. $C_1$ has 500 turns and radius of 1 cm. $C_2$ has 200 turns and radius of 20 cm. $C_2$ carries a time dependent current $I(t) = (5t^2 - 2t + 3)A$, where $t$ is in secon(d) The emf induced in $C_1$ (in mV), at the instant $t = 1s$ is $\frac{4}{x}$. The value of $x$ is ...... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at two concentric circular coils lying flat on the $xy$-plane. The outer coil, $C_2$, is quite large with a radius of $20 	ext{ cm}$ and $200$ turns. It carries a time-varying current given by $I(t) = 5t^2 - 2t + 3$.

Nestled right at its center is the much smaller inner coil, $C_1$, with a radius of just $1 	ext{ cm}$ but packed tightly with $500$ turns. Our mission is to find the induced electromotive force (EMF) in this inner coil at the exact moment $t = 1 	ext{ s}$.

Because the inner coil is so small compared to the outer one ($r_1 \ll r_2$), we can safely assume that the magnetic field produced by the outer coil is practically uniform over the entire area of the inner coil. This is a crucial approximation that saves us from a nightmare of complex integration!

### The Master Equation

The magnetic field at the center of the outer coil $C_2$ is given by the standard formula:
$$B_2 = \frac{\mu_0 N_2 I(t)}{2 r_2}$$

This magnetic field pierces through the inner coil $C_1$, creating a magnetic flux. The total flux linked with all $N_1$ turns of the inner coil is:
$$\Phi_1 = N_1 B_2 A_1 = N_1 \left( \frac{\mu_0 N_2 I(t)}{2 r_2} ight) (\pi r_1^2)$$

According to **Faraday's Law of Electromagnetic Induction**, the magnitude of the induced EMF is the rate of change of this magnetic flux. Since everything else is constant, the only thing changing with time is the current $I(t)$:
$$e = \left| \frac{d\Phi_1}{dt} ight| = \frac{\mu_0 N_1 N_2 \pi r_1^2}{2 r_2} \frac{dI}{dt}$$

### Final Calculation

First, let's find the rate of change of current at $t = 1 	ext{ s}$. Differentiating the given current equation:
$$\frac{dI}{dt} = \frac{d}{dt}(5t^2 - 2t + 3) = 10t - 2$$

At $t = 1 	ext{ s}$, this becomes:
$$\frac{dI}{dt} = 10(1) - 2 = 8 	ext{ A/s}$$

Now, we substitute all our known values into the EMF equation. **Watch out for the units!** Radii must be in meters: $r_1 = 10^{-2} 	ext{ m}$ and $r_2 = 0.2 	ext{ m}$.
$$e = \frac{(4\pi 	imes 10^{-7}) 	imes 500 	imes 200 	imes \pi 	imes (10^{-2})^2}{2 	imes 0.2} 	imes 8$$

Let's simplify the numerator:
$$e = \frac{4\pi^2 	imes 10^{-7} 	imes 10^5 	imes 10^{-4}}{0.4} 	imes 8$$
$$e = \frac{32\pi^2 	imes 10^{-6}}{0.4} = 80\pi^2 	imes 10^{-6} 	ext{ V}$$

Here is a classic JEE trick: approximate $\pi^2 \approx 10$.
$$e \approx 80 	imes 10 	imes 10^{-6} = 800 	imes 10^{-6} 	ext{ V} = 0.8 	ext{ mV}$$

The problem states that the induced EMF is $\frac{4}{x} 	ext{ mV}$. Equating the two:
$$0.8 = \frac{4}{x} \implies x = \frac{4}{0.8} = 5$$

And there we have it! The value of $x$ is **$5$**.

Enter Numerical Value:

Visualized Solution

$C_{1}$ and $C_{2}$ Parameters$

$B$ -Field due to Outer Coil$

Magnetic Flux through Inner Coil$

Faraday's Law of Induction$

Rate of Change of Current$

Substituting the Values$

Calculating the EMF$

Finding $x$

The Way Forward$

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

The flux linked with a coil at any instant $t$ is given by $ϕ = 10 t^{2} - 50 t + 250$ . The induced emf at $t = 3 s$ is

At time $t = 0$ magnetic field of 1000 gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to 500 gauss, in the next 5 s, then induced emf in the loop is

A coil having $n$ turns and resistance $R Ω$ is connected with a galvanometer of resistance $4 R Ω$ . This combination is moved for time $t$ seconds from a magnetic field $W_{1}$ weber to $W_{2}$ weber. The induced current in the circuit is

In a coil of resistance $100 Ω$ , a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is

A planar loop of wire rotates in a uniform magnetic field. Initially at $t = 0$ , the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of $10 s$ about an axis in its plane, then the magnitude of induced emf will be maximum and minimum respectively at

In the given figure, the magnetic flux through the loop increases according to the relation $ϕ_{B} (t) = 10 t^{2} + 20 t$ , where $ϕ_{B}$ is in milliwebers and $t$ is in seconds. The magnitude of current through $R = 2Ω$ resistor at $t = 5$ s is ...... mA.