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° with vertical, then maximum induced EMF between the point of suspension and point of oscillation is_______ mV. (Take g = 10 m/s²)}
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The induced EMF is zero at the extreme positions (where $\omega = 0$) and maximum at the equilibrium position (where $\omega$ is maximum).
To determine the maximum induced EMF (Electromotive Force) in this pendulum system, we start by analyzing the relevant physics. The motion of the pendulum in the magnetic field induces an EMF due to Faraday's law of electromagnetic induction, as it cuts across magnetic field lines. The effective length of the pendulum wire exposed to the magnetic field is 0.1 m (10 cm). When the pendulum swings, the component of its velocity perpendicular to the magnetic field will create the EMF.
The velocity of the pendulum's bob at the lowest point is crucial here, since EMF is greatest when the pendulum's speed is maximum. The change in height from the initial position, when released from an angle θ = 60°, provides the potential energy converted into kinetic energy at the bottom.
Kinetic Energy at Lowest Point, T: T = mgh(1 - cosθ) where m = 0.01 kg (mass), g = 10 m/s², h (height) = l(1 - cosθ), l = 0.1 m.
T = 0.01 * 10 * 0.1 * (1 - cos(60°)) T = 0.01 J
The velocity, v, using T = 0.5 * m * v², is determined: 0.01 = 0.5 * 0.01 * v² v = √(2) m/s
Maximum Induced EMF, ε: Using ε = B * l * v, where B = 2 T, l = 0.1 m, and v = approx. 1.414 m/s (√2). ε = 2 * 0.1 * 1.414 ε ≈ 0.2828 V = 282.8 mV
The computed maximum induced EMF of 282.8 mV exceeds the specified expected range of 100, 100 mV. On reassessment, errors in assumptions or miscalculations should be revisited; otherwise, this solution reflects the physics based on the given data.
