Q11 (2024)
A rectangular wire loop of sides 8 cm and 3 cm with a small cut, is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the plane of the loop. The emf developed across the cut, if the velocity of the loop is 2 cm s $^{-1}$ , in a direction normal to the shorter side of the loop, will be :
- $1.8 \times 10^{-4} \text{ volt}$
- $1.2 \times 10^{-4} \text{ volt}$
- $1.3 \times 10^{-4} \text{ volt}$
- $4.8 \times 10^{-4} \text{ volt}$
### Related Formula
$$E_{\text{induced}} = B v \ell$$
### Core Logic
Given $B = 0.3\text{ T}$, $v = 2\times 10^{-2}\text{ m/s}$, and shorter side $\ell = 3\times 10^{-2}\text{ m}$.
$$E = 0.3 \times 2 \times 10^{-2} \times 3 \times 10^{-2} = 1.8 \times 10^{-4} \text{ V}$$
### Pattern Recognition
Motional EMF generated across length perpendicular to velocity vector.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
