Given: \[ \tau = mB \] \[ K_{\theta} = \frac{IA \cdot NB}{I \cdot \theta} = \frac{A \cdot N \cdot B}{K} \quad \text{(1)} \] \[ \left( \frac{\theta}{\theta_1} \right) = \frac{I_1}{I_2} = \frac{A \cdot N \cdot B}{K} \quad \text{(2)} \] \[ \Rightarrow \lambda = a \cdot N \quad \text{(3)} \] \[ V_{\theta} = I \cdot R = \text{Current Sensitivity} \quad \text{(4)} \] \[ \text{Voltage sensitivity} \propto \text{Current sensitivity} \]
Correct Answer: (A)
The formulae involve relationships for current and voltage sensitivities. Equation (1) gives the relationship between the magnetic field \( B \) and the area of the coil \( A \). From Equation (2), we can deduce that voltage sensitivity is directly proportional to current sensitivity. This shows a clear relation where \( R \) remains constant.
A circular coil of diameter 15 mm having 300 turns is placed in a magnetic field of 30 mT such that the plane of the coil is perpendicular to the direction of the magnetic field. The magnetic field is reduced uniformly to zero in 20 ms and again increased uniformly to 30 mT in 40 ms. If the EMFs induced in the two time intervals are \( e_1 \) and \( e_2 \) respectively, then the value of \( e_1 / e_2 \) is:
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.} 

\[ f(x) = \left\{ \begin{array}{ll} 1 - 2x & \text{if } x < -1 \\ \frac{1}{3}(7 + 2|x|) & \text{if } -1 \leq x \leq 2 \\ \frac{11}{18} (x-4)(x-5) & \text{if } x > 2 \end{array} \right. \]