Q194 / 240 · KCET 2026 · Physics
In the figure shown, the conductor PQ of length $l$ is moved from $x = 0$ to $x = b$ and then up to $x = 2b$ with a constant velocity $\vec{v}$. A uniform magnetic field $\vec{B}$ is perpendicular to the plane of the paper and extends from $x = 0$ to $x = b$ and it is zero from $x > b$. The magnitude of emf Induced in the conductor is
- A$B l x; 0 \leq x \leq b$
- BZero; $0 \leq x \leq b$
- Correct answerC$B l v; 0 \leq x \leq b$
- D$B l v; b \leq x \leq 2b$
Solution
When the conductor PQ moves from $x = 0$ to $x = b$, it moves through a uniform magnetic field $\vec{B}$ which is perpendicular to both its length $l$ and velocity $\vec{v}$.
The motional emf induced across the ends of the conductor is given by the formula:
$E = B l v$
Since $B$, $l$, and $v$ are constant, the induced emf is constant and equal to $B l v$ in the region $0 \leq x \leq b$.
When the conductor moves from $x = b$ to $x = 2b$, the magnetic field is zero ($B = 0$).
Therefore, the induced emf in this region is zero.
Thus, the magnitude of emf induced is $B l v$ for $0 \leq x \leq b$.
Answer: $B l v; 0 \leq x \leq b$
The motional emf induced across the ends of the conductor is given by the formula:
$E = B l v$
Since $B$, $l$, and $v$ are constant, the induced emf is constant and equal to $B l v$ in the region $0 \leq x \leq b$.
When the conductor moves from $x = b$ to $x = 2b$, the magnetic field is zero ($B = 0$).
Therefore, the induced emf in this region is zero.
Thus, the magnitude of emf induced is $B l v$ for $0 \leq x \leq b$.
Answer: $B l v; 0 \leq x \leq b$
Hence, the correct answer is option: 'C'.