As \( \varepsilon \big|_{t=0.5 \, \text{sec}} = -\frac{d\phi}{dt} \) \[ = - A \frac{dB}{dt} \quad [\because \theta = 0^\circ \Rightarrow \cos \theta = 1] \] \[ = - \pi \times \left( \frac{10}{\sqrt{\pi}} \right)^2 \times 10^{-4} \times \frac{0 - 0.5}{0.5} = 10^{-2}V = 10 \, \text{mV} \] As \( \frac{dB}{dt} = \) constant \(\Rightarrow\) Induced emf will not change with time. So, \[ \varepsilon \big|_{0.5 \, \text{sec}} = \varepsilon \big|_{0.25 \, \text{sec}} = 10 \, \text{mV} \]
A thin transparent film with refractive index 1.4 is held on a circular ring of radius 1.8 cm. The fluid in the film evaporates such that transmission through the film at wavelength 560 nm goes to a minimum every 12 seconds. Assuming that the film is flat on its two sides, the rate of evaporation is:

An infinite wire has a circular bend of radius \( a \), and carrying a current \( I \) as shown in the figure. The magnitude of the magnetic field at the origin \( O \) of the arc is given by: