Step 1: Use Faraday's law.
Induced emf = $B l v$ for a rod of length $l$ moving with velocity $v$ perpendicular to a magnetic field.
Here the side of length $w$ cuts the magnetic flux, so magnitude = $B w v$.
Step 2: Analyse entry into field.
When the front of the loop enters the field region, only one vertical segment is cutting flux ⇒ emf = $+Bwv$.
Step 3: When the loop is fully inside.
Two opposite vertical sides cut equal flux in opposite directions ⇒ net emf = $0$.
Step 4: When the loop leaves the field.
Only the back segment cuts the flux ⇒ emf = $-Bwv$.
Step 5: Conclusion.
The emf–position graph must show:
• +Bwv at entry
• 0 while fully inside
• –Bwv at exit
This matches option (B).

In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:
A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of \( 2 \times 10^5 \, \text{m/s} \). When the electric field is switched off, the proton moves along a circular path of radius 2 cm. The magnitude of electric field is \( x \times 10^4 \, \text{N/C} \). The value of \( x \) is \(\_\_\_\_\_\). (Take the mass of the proton as \( 1.6 \times 10^{-27} \, \text{kg} \)).
Due to presence of an em-wave whose electric component is given by \( E = 100 \sin(\omega t - kx) \, NC^{-1} \), a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as:
