Consider a small element of length dx at a distance x from the centre of the circle described by the rod. As this element moves perpendicular to the field with a linear velocity v = xω, the emf developed in the element dx is dε = Bvdx = B(xω)dx
This rod is made up of many such elements, moving perpendicular to the field. The emf developed across two ends is
\(\epsilon =\int { d\epsilon } =\int _{ 0 }^{ l }{ B\omega xdx } =B\omega { { \left[ \frac { { x }^{ 2 } }{ 2 } \right] } }_{ 0 }^{ l }\)
\(\epsilon =\frac { 1 }{ 2 } B\omega { l }^{ 2 }\)

\(\Phi_B\) = (2t³ + 3t² + 8t + 5)mWb; N = 1;t = 3 s
i) \(ε=\frac { d(N{ \Phi }_{ B }) }{ dt } \)
\(=\frac { d }{ t } \left( { 2t }^{ 3 }+{ 3t }^{ 2 }+8t+5 \right) \times { 10 }^{ -3 }\)
= (6t² + 6t + 8) x 10^-3 V
At t = 3 s,
ε = [( 6 x 9) + (6 x 3) + 8] x 10^-3
= 80 x 10^-3V = 80mV
(ii) As time passes, the magnetic flux linked with the loop increases. According to Lenz’s law, the direction of the induced current should be in a way so as to oppose the flux increase. So, the induced current flows in such a way to produce a magnetic field opposite to the given field. This magnetic field is perpendicularly outwards. Therefore, the induced current flows in anticlockwise direction.

Yes! An emf will be induced in the wire because it moves perpendicular to the horizontal component of Earth’s magnetic field and hence it cuts the magnetic lines of Earth's magnetic field.

The magnetic field of a cylindrical magnet is symmetrical about its axis. As the magnet is rotated along the axis of the solenoid, there is no induced current in the solenoid because the flux linked with the solenoid does not change due to the rotation of the magnet.

B = 4.1 x 10^–5 T; θ = 90^o – 47^o = 43° ;
A = 3m²
We know that   \(\Phi_{B}=B A \cos \theta\) 
\(\Phi_{\mathrm{B}}\) = 4.1 x 10^–5 x 3 x cos 43^o  
= 4.1 x 10^–5 x 3 x 0.7314
= 89.96 \(\mu \mathrm{Wb}\).