Q46 (2026)
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 pendulum is released from an angle of $60^{\circ}$ with vertical, then maximum induced EMF between the point of suspension and point of oscillation is ____ mV. (Take $g = 10 \text{ m/s}^{2}$)
### Related Formula
$$\varepsilon = \frac{1}{2} B\omega l^{2}$$
$$mgl(1 - \cos \theta) = \frac{1}{2}(ml^{2})\omega^{2}$$
### Core Logic
The metallic string of the pendulum cuts the perpendicular magnetic field lines as it swings. The induced EMF along the length of a rotating rod is maximum when its angular velocity $\omega$ is maximum. This maximum $\omega$ occurs at the lowest point of the swing.
### Step 1: Find Maximum Angular Velocity
Using conservation of mechanical energy from extreme to mean position:
$$mgl(1 - \cos 60^{\circ}) = \frac{1}{2} I \omega_{\max}^{2}$$
$$mgl\left(1 - \frac{1}{2}\right) = \frac{1}{2}(ml^{2})\omega_{\max}^{2}$$
$$\frac{1}{2}mgl = \frac{1}{2}ml^{2}\omega_{\max}^{2}$$
$$\omega_{\max} = \sqrt{\frac{g}{l}}$$
### Step 2: Substitute Values
Given $l = 10 \text{ cm} = 0.1 \text{ m}$ and $g = 10 \text{ m/s}^{2}$.
$$\omega_{\max} = \sqrt{\frac{10}{0.1}} = \sqrt{100} = 10 \text{ rad/s}$$
### Step 3: Calculate Maximum EMF
$$\varepsilon_{\max} = \frac{1}{2} B \omega_{\max} l^{2}$$
$$\varepsilon_{\max} = \frac{1}{2} (2 \text{ T})(10 \text{ rad/s})(0.1 \text{ m})^{2}$$
$$\varepsilon_{\max} = 10 \times 0.01 = 0.1 \text{ V}$$
Convert to millivolts:
$$\varepsilon_{\max} = 100 \text{ mV}$$
### Pattern Recognition
Sees: "pendulum wire in magnetic field" → It's just a rotating rod! EMF is $\frac{1}{2} B \omega l^2$. Find max $\omega$ via energy conservation at the lowest point.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Class 11 Physics: Work, Energy and Power
