Q178 / 200 · MHT CET - Medical 2026 (21 April Shift 1) · Physics
A metallic rod of length $'l'$ is tied to a string of length $'2l'$ and made to rotate with angular speed $'2\omega'$ on a horizontal table with one end of the string fixed. If there exists a vertical magnetic field 'B' in the region, the emf $'e'$ induced across the ends of rod is
- A$5B\omega l^3$
- B$\dfrac{3}{2}B\omega l^3$
- C$2\ B\omega l^2$
- Correct answerD$5\,B\omega\,l^2$
Solution
Let the fixed end of the string be the origin. The rod is situated between the distances $r_1 = 2l$ and $r_2 = 2l + l = 3l$ from the fixed end.
Consider a small element of the rod of length $dx$ at a distance $x$ from the fixed end. The velocity of this element is $v = x(2\omega)$.
The motional emf induced in this small element is $de = Bv\,dx = B(x \cdot 2\omega)\,dx$.
The total induced emf across the ends of the rod is obtained by integrating $de$ from $x = 2l$ to $x = 3l$:
$e = \int_{2l}^{3l} 2B\omega x\,dx$
$e = 2B\omega \left[ \dfrac{x^2}{2} \right]_{2l}^{3l}$
$e = B\omega ((3l)^2 - (2l)^2)$
$e = B\omega (9l^2 - 4l^2)$
$e = 5B\omega l^2$
Answer: $5\,B\omega\,l^2$
Consider a small element of the rod of length $dx$ at a distance $x$ from the fixed end. The velocity of this element is $v = x(2\omega)$.
The motional emf induced in this small element is $de = Bv\,dx = B(x \cdot 2\omega)\,dx$.
The total induced emf across the ends of the rod is obtained by integrating $de$ from $x = 2l$ to $x = 3l$:
$e = \int_{2l}^{3l} 2B\omega x\,dx$
$e = 2B\omega \left[ \dfrac{x^2}{2} \right]_{2l}^{3l}$
$e = B\omega ((3l)^2 - (2l)^2)$
$e = B\omega (9l^2 - 4l^2)$
$e = 5B\omega l^2$
Answer: $5\,B\omega\,l^2$
Hence, the correct answer is option: 'D'.