CBSE 2025, Set 2

Question

State Lenz's law. A rod MN of length $L$ is rotated about an axis passing through its end M perpendicular to its length, with a constant angular velocity $\omega$ in a uniform magnetic field $\vec{B}$ parallel to the axis. Obtain an expression for emf induced between its ends.

Edvaya · Accepted Answer

The correct answer is Subjective. **Lenz's Law:**
The polarity of the induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it.

**Expression for induced emf:**

![Rotating Rod in Magnetic Field](https://www.edvaya.com/images/cbse/physics/2025/set2/q23a-rod-rotating-magnetic-field-answer.webp)

Consider a small element of the rod of length $dr$ at a distance $r$ from the axis of rotation $M$. The linear velocity of this element is $v = r\omega$.
The magnitude of the emf generated across the length $dr$ of the rod as it moves at right angles to the magnetic field is given by:
$$d\epsilon = B v \, dr$$
$$d\epsilon = B (r\omega) \, dr$$

The total emf $\epsilon$ induced across the ends of the rod is the integral of $d\epsilon$ from $r = 0$ to $r = L$:
$$\epsilon = \int d\epsilon = \int_{0}^{L} B \omega r \, dr$$
$$\epsilon = B \omega \left[ \frac{r^2}{2} ight]_{0}^{L}$$
$$\boxed{\epsilon = \frac{1}{2} B L^2 \omega}$$

**Alternatively (using area swept):**
In time $dt$, the angle swept by the rod is $d	heta = \omega dt$.
Area swept by the rod is $dA = \frac{1}{2} L^2 d	heta$.
Induced emf $\epsilon = B 	imes \frac{dA}{dt} = B 	imes \frac{d}{dt} \left( \frac{1}{2} L^2 	heta ight) = \frac{1}{2} B L^2 \frac{d	heta}{dt} = \frac{1}{2} B L^2 \omega$.

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Physics > Magnetism Induction And Alternating Current > Electromagnetic Induction

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