A rectangular wire loop of sides 8 ~cm$8 \mathrm{~cm}$ and 3 ~cm$3 \mathrm{~cm}$ with a small cut, is moving out of a region of uniform magnetic field of magnitude 0.3 ~T$0.3 \mathrm{~T}$ 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$2 \mathrm{~cm \ s^{-1}}$, in a direction normal to the shorter side of the loop, will be :

Related Formula

Core Logic

Since the loop is moving in a direction normal to the shorter side (3 ~cm$3 \mathrm{~cm}$), it means the longer side (8 ~cm$8 \mathrm{~cm}$) is the one moving across the magnetic field lines. Wait, if it moves normal to the shorter side, it is sliding along the longer side, so the shorter side is 'cutting' the flux lines.

Let's re-read carefully: "in a direction normal to the shorter side of the loop". This means the velocity vector is perpendicular to the 3 ~cm$3 \mathrm{~cm}$ side. Thus, the effective length l$l$ that cuts the magnetic flux is the shorter side, l = 3 ~cm$l = 3 \mathrm{~cm}$.

Step 1: Calculate Induced EMF

Given: B = 0.3 ~T$B = 0.3 \mathrm{~T}$ v = 2 ~cm/s = 2 × 10⁻² ~m/s$v = 2 \mathrm{~cm/s} = 2 \times 10^{-2} \mathrm{~m/s}$ l = 3 ~cm = 3 × 10⁻² ~m$l = 3 \mathrm{~cm} = 3 \times 10^{-2} \mathrm{~m}$ (since velocity is normal to it)

Pattern Recognition

In motional emf, ε = Bvl$\varepsilon = Bvl$, the length l$l$ must be mutually perpendicular to both the velocity v$\vec{v}$ and the magnetic field B$\vec{B}$. "Normal to the shorter side" means the shorter side acts as the active length.

Chapter Mix

Class 12 Physics: Electromagnetic Induction