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For a place the Earth's magnetic field i...

For a place the Earth's magnetic field is `4 xx 10^(-5) T` and angle of dip is `60^(@)`. A wheel of radius 60 cm having 10 spokes is rotating with a speed of `150 "rev" // "min"` in plane perpendicular to Earth's horizontal magnetic field. Find the induced emf between axle and rim of the wheel.

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To find the induced EMF between the axle and rim of the rotating wheel, we can follow these steps: ### Step 1: Determine the Horizontal Component of Earth's Magnetic Field Given: - Total Earth's magnetic field, \( B = 4 \times 10^{-5} \, \text{T} \) - Angle of dip, \( \delta = 60^\circ \) The horizontal component of the magnetic field \( B_H \) can be calculated using the formula: \[ B_H = B \cos(\delta) \] Substituting the values: \[ B_H = 4 \times 10^{-5} \cos(60^\circ) = 4 \times 10^{-5} \times \frac{1}{2} = 2 \times 10^{-5} \, \text{T} \] ### Step 2: Convert the Rotational Speed to Radians per Second Given: - Rotational speed = 150 revolutions per minute (rev/min) To convert this to radians per second: \[ \omega = 150 \, \text{rev/min} \times \frac{2\pi \, \text{rad}}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{s}} = 5\pi \, \text{rad/s} \] ### Step 3: Calculate the Length of the Spoke Given: - Radius of the wheel \( r = 60 \, \text{cm} = 0.6 \, \text{m} \) The length of one spoke \( L \) is equal to the radius: \[ L = 0.6 \, \text{m} \] ### Step 4: Use the Induced EMF Formula The formula for the induced EMF \( E \) in a rotating wheel is given by: \[ E = \frac{1}{2} B_H L^2 \omega \] Substituting the values: \[ E = \frac{1}{2} \times (2 \times 10^{-5}) \times (0.6)^2 \times (5\pi) \] Calculating: \[ E = \frac{1}{2} \times 2 \times 10^{-5} \times 0.36 \times 5\pi \] \[ E = 10^{-5} \times 0.18 \times 5\pi \] \[ E \approx 10^{-5} \times 0.18 \times 15.70796 \approx 2.82 \times 10^{-4} \, \text{V} \] ### Step 5: Final Calculation Calculating the final value: \[ E \approx 5.65 \times 10^{-5} \, \text{V} \] ### Conclusion The induced EMF between the axle and rim of the wheel is approximately: \[ E \approx 5.65 \times 10^{-5} \, \text{V} \]
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