Physics Electromagnetic Induction 5 Marks
Appeared in: 2024 (3 sets)
OR (b) (i) A rectangular coil of N turns and area of cross-section A is rotated at a steady angular speed $\omega$ in a uniform magnetic field. Obtain an expression for the emf induced in the coil at any instant of time. (ii) Two coplanar and concentric circular loops $L_1$ and $L_2$ are placed coaxially with their centres coinciding. The radii of $L_1$ and $L_2$ are 1 cm and 100 cm respectively. Calculate the mutual inductance of the loops. (Take $\pi^2 = 10$)
✓ Verified CBSE Marking Scheme Official Evaluation Steps & EL10 Insight

Part (i): Expression for Induced EMF in a Rotating Coil

Step 1 (Governing Law & Formula): [1 Mark]

The magnetic flux $\phi$ linked with a coil of $N$ turns, area $A$, in a magnetic field $B$ at an angle $\theta$ is given by:

$$\phi = N B A \cos(\theta)$$

Since the coil rotates with angular speed $\omega$, $\theta = \omega t$. Thus, $\phi = N B A \cos(\omega t)$. According to Faraday’s Law of Induction, the induced emf $\varepsilon$ is:

$$\varepsilon = -\frac{d\phi}{dt}$$

Step 2 (Mathematical Derivation): [1.5 Marks]

Substituting the expression for flux into Faraday's law:

$$\varepsilon = -\frac{d}{dt} (N B A \cos(\omega t))$$

Since $N, B,$ and $A$ are constants:

$$\varepsilon = -N B A \frac{d}{dt} (\cos(\omega t))$$

Using the derivative $\frac{d}{dt}(\cos(\omega t)) = -\omega \sin(\omega t)$:

$$\varepsilon = -N B A (-\omega \sin(\omega t))$$

Step 3 (Final Expression): [0.5 Mark]

$$\varepsilon = N B A \omega \sin(\omega t)$$

Where $\varepsilon_0 = N B A \omega$ is the peak value of the induced emf.


Part (ii): Calculation of Mutual Inductance

Step 1 (Formula & Given Data): [0.5 Mark]

Given: $r_1 = 1 \text{ cm} = 10^{-2} \text{ m}$, $r_2 = 100 \text{ cm} = 1 \text{ m}$.

The mutual inductance $M$ for two concentric coplanar loops (where $r_1 \ll r_2$) is derived from the flux through the smaller loop due to the magnetic field of the larger loop:

$$M = \frac{\mu_0 \pi r_1^2}{2 r_2}$$

Step 2 (Mathematical Working): [1 Mark]

Substitute the values ($\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$ and $\pi^2 = 10$):

$$M = \frac{(4\pi \times 10^{-7}) \times \pi \times (10^{-2})^2}{2 \times 1}$$ $$M = \frac{4 \pi^2 \times 10^{-7} \times 10^{-4}}{2}$$ $$M = 2 \pi^2 \times 10^{-11}$$

Step 3 (Final Answer): [0.5 Mark]

Using $\pi^2 = 10$:

$$M = 2 \times 10 \times 10^{-11} = 2 \times 10^{-10} \text{ H}$$

💡 EL10 Concept Insight:

Think of the rotating coil like a spinning door in a breeze; as it turns, it "catches" different amounts of magnetic field lines, which forces electrons to move and creates electricity. For the loops, the big loop acts like a magnet, and the tiny loop inside just "listens" to the magnetic field created by