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 s¹, in a direction normal to the shorter side of the loop, will be: ____.

**Step 1 – Identify the relevant law**
Faraday’s law states that the magnitude of the induced emf (𝜀) around a moving conductor equals the rate of change of magnetic flux through the loop:

$𝜀\; =\; B\backslash ,\backslash frac\{dA\}\{dt\}$

where B is the magnetic field strength and A is the portion of the loop still inside the field.

**Step 2 – Relate the change of area to the motion**
The loop moves outward in a direction normal to the shorter side (3 cm). Consequently, the side that is parallel to the motion is the longer side (8 cm). As the loop exits the field, the area inside the field decreases at a rate equal to the product of the length of the side parallel to the motion (ℓ) and the velocity (v):

$\backslash frac\{dA\}\{dt\}=\; -\backslash ,\backslash ell\backslash ,v$

The negative sign indicates a decreasing area, but the emf magnitude uses the absolute value.

**Step 3 – Insert the given numerical values**
Convert all quantities to SI units:

• B = 0.3 T
• ℓ = 8 cm = 0.08 m
• v = 2 cm s⁻¹ = 0.02 m s⁻¹

**Step 4 – Compute the emf**

$𝜀\; =\; B\backslash ,\backslash ell\backslash ,v\; =\; (0.3\backslash \; \backslash text\{T\})\backslash ,(0.08\backslash \; \backslash text\{m\})\backslash ,(0.02\backslash \; \backslash text\{m s\}^\{-1\})$

$𝜀\; =\; 0.3\; \backslash times\; 0.0016\backslash \; \backslash text\{V\}$

$𝜀\; =\; 0.00048\backslash \; \backslash text\{V\}$

Expressed in scientific notation:

$𝜀\; =\; 4.8\; \backslash times\; 10^\{-4\}\backslash \; \backslash text\{V\}$

**Conclusion**
The emf developed across the cut while the rectangular loop leaves the magnetic field is therefore:

$𝜀\; =\; 4.8\; \backslash times\; 10^\{-4\}\backslash \; \backslash text\{volt\}$