A simple pendulum made of mass 10 g and a metallic wire of length 10 cm is suspended vertically in a uniform magnetic field of 2 T. The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulwn is released from an angle of 60° with vertical, then maximum induced EMF between the point of s uspension and point of oscillation is ______mV (Take $$g= 10 m/s^{2}$$)

The pendulum consists of a metallic wire of length $$L = 10 \text{ cm} = 0.1 \text{ m}$$ and a bob of mass $$m = 10 \text{ g} = 0.01 \text{ kg}$$, oscillating in a uniform magnetic field $$B = 2 \text{ T}$$ perpendicular to the plane of oscillation. The maximum induced EMF occurs when the angular velocity is maximum, which is at the lowest point of the swing.

The induced EMF in a rotating conductor of length $$L$$ about one end in a uniform magnetic field $$B$$ perpendicular to the plane of rotation is given by:

$$\mathcal{E} = \frac{1}{2} B \omega L^2$$

where $$\omega$$ is the angular velocity.

To find the maximum angular velocity $$\omega_{\text{max}}$$, use conservation of energy. The initial angle is $$\theta_0 = 60^\circ$$. The height $$h$$ above the lowest point is:

$$h = L (1 - \cos \theta_0)$$

Substitute $$\cos 60^\circ = 0.5$$:

$$h = 0.1 \times (1 - 0.5) = 0.1 \times 0.5 = 0.05 \text{ m}$$

The potential energy at the initial position converts to kinetic energy at the lowest point:

$$m g h = \frac{1}{2} m v_{\text{max}}^2$$

where $$g = 10 \text{ m/s}^2$$. Solve for $$v_{\text{max}}$$:

$$v_{\text{max}} = \sqrt{2 g h} = \sqrt{2 \times 10 \times 0.05} = \sqrt{1} = 1 \text{ m/s}$$

The maximum angular velocity $$\omega_{\text{max}}$$ is related to $$v_{\text{max}}$$ by:

$$\omega_{\text{max}} = \frac{v_{\text{max}}}{L} = \frac{1}{0.1} = 10 \text{ rad/s}$$

Now substitute into the EMF formula:

$$\mathcal{E}_{\text{max}} = \frac{1}{2} B \omega_{\text{max}} L^2 = \frac{1}{2} \times 2 \times 10 \times (0.1)^2$$

Calculate step by step:

$$(0.1)^2 = 0.01$$

$$\mathcal{E}_{\text{max}} = \frac{1}{2} \times 2 \times 10 \times 0.01 = 1 \times 10 \times 0.01 = 0.1 \text{ V}$$

Convert to millivolts: $$0.1 \text{ V} = 100 \text{ mV}$$.

Thus, the maximum induced EMF is $$100 \text{ mV}$$.