The induced charge in the coil is given by Faraday's law: \[ Q = \frac{N \cdot A \cdot B}{R} \cdot \Delta t \] where:
\( N = 100 \) is the number of turns,
\( A = 0.05 \, \text{m}^2 \) is the area of each turn,
\( B = 90 \, \text{mT} = 0.09 \, \text{T} \) is the magnetic field strength,
\( R = 1.5 \, \Omega \) is the total resistance of the coil.
Substitute these values into the formula: \[ Q = \frac{100 \cdot 0.05 \cdot 0.09}{1.5} = 0.30 \, \text{C} \] Thus, the correct answer is \( 0.30 \, \text{C} \).
A coil of area A and N turns is rotating with angular velocity \( \omega\) in a uniform magnetic field \(\vec{B}\) about an axis perpendicular to \( \vec{B}\) Magnetic flux \(\varphi \text{ and induced emf } \varepsilon \text{ across it, at an instant when } \vec{B} \text{ is parallel to the plane of the coil, are:}\)
Conductor wire ABCDE with each arm 10 cm in length is placed in magnetic field of $\frac{1}{\sqrt{2}}$ Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of $10 \mathrm{~cm} / \mathrm{s}$, induced emf between points A and E is _______ mV.} 