Q163 / 180 · COMEDK 2024 (Afternoon Shift) · Physics
A metallic rod of length '$a$' is rotated with an angular frequency of $0.2 \mathrm{~rads}^{-1}$ about an axis normal to the rod passing through its one end. A constant and uniform magnetic field of '$\mathrm{B}$' T parallel to the axis exists everywhere. The emf developed across the ends of the rod is
- Correct answerA$\dfrac{\mathrm{Ba}^2}{10}$
- B$\dfrac{\mathrm{Ba}^2}{2}$
- C$\dfrac{\mathrm{Ba}^2}{50}$
- D$\dfrac{\mathrm{Ba}^2}{5}$
Solution
The motional electromotive force (emf) induced in a rod of length $a$ rotating with angular velocity $\omega$ in a uniform magnetic field $B$ perpendicular to the plane of rotation is given by the integral of $d\epsilon = B v dr$, where $v = \omega r$.
The emf $E$ is calculated as:
$E = \int_{0}^{a} B \omega r dr$
$E = B \omega \left[ \dfrac{r^2}{2} \right]_{0}^{a}$
$E = \dfrac{1}{2} B \omega a^2$
Given the angular frequency $\omega = 0.2 \text{ rad s}^{-1}$, substitute this value into the expression:
$E = \dfrac{1}{2} \times B \times 0.2 \times a^2$
$E = 0.1 \times B \times a^2$
$E = \dfrac{1}{10} B a^2 = \dfrac{B a^2}{10}$
Answer: $\dfrac{\mathrm{Ba}^2}{10}$
The emf $E$ is calculated as:
$E = \int_{0}^{a} B \omega r dr$
$E = B \omega \left[ \dfrac{r^2}{2} \right]_{0}^{a}$
$E = \dfrac{1}{2} B \omega a^2$
Given the angular frequency $\omega = 0.2 \text{ rad s}^{-1}$, substitute this value into the expression:
$E = \dfrac{1}{2} \times B \times 0.2 \times a^2$
$E = 0.1 \times B \times a^2$
$E = \dfrac{1}{10} B a^2 = \dfrac{B a^2}{10}$
Answer: $\dfrac{\mathrm{Ba}^2}{10}$
Hence, the correct answer is option: 'A'.