Q176 / 180 · COMEDK 2024 (Morning Shift) · Physics
A metallic rod of $2 \mathrm{~m}$ length is rotated with a frequency $100 \mathrm{~Hz}$ about an axis passing through the centre of the circular ring of radius $2 \mathrm{~m}$. A constant magnetic field $2 \mathrm{~T}$ is applied parallel to the axis and perpendicular to the length of the rod. The emf developed across the ends of the rod is :
- Correct answerA800 $\pi$ volt
- B400 $\pi$ volt
- C1600 $\pi$ volt
- D1600 volt
Solution
The induced emf $e$ across a rod of length $l$ rotating with angular velocity $\omega$ in a uniform magnetic field $B$ perpendicular to the plane of rotation is given by the formula $e = \dfrac{1}{2} B \omega l^2$.
Given parameters are:
Length of the rod $l = 2 \text{ m}$
Frequency $f = 100 \text{ Hz}$
Magnetic field $B = 2 \text{ T}$
The angular velocity $\omega$ is calculated as $\omega = 2 \pi f = 2 \pi \times 100 = 200 \pi \text{ rad/s}$.
Substituting these values into the formula:
$e = \dfrac{1}{2} \times 2 \times (200 \pi) \times (2)^2$
$e = 1 \times 200 \pi \times 4$
$e = 800 \pi \text{ V}$.
Answer: 800 $\pi$ volt
Given parameters are:
Length of the rod $l = 2 \text{ m}$
Frequency $f = 100 \text{ Hz}$
Magnetic field $B = 2 \text{ T}$
The angular velocity $\omega$ is calculated as $\omega = 2 \pi f = 2 \pi \times 100 = 200 \pi \text{ rad/s}$.
Substituting these values into the formula:
$e = \dfrac{1}{2} \times 2 \times (200 \pi) \times (2)^2$
$e = 1 \times 200 \pi \times 4$
$e = 800 \pi \text{ V}$.
Answer: 800 $\pi$ volt
Hence, the correct answer is option: 'A'.