Suppose you wanted to use Earth's magnetic field to make an ac generator at a location where the magnitude of the field is \(0.50 \mathrm{mT}\). Your coil has 1000.0 turns and a radius of \(5.0 \mathrm{cm} .\) At what angular velocity would you have to rotate it in order to generate an emf of amplitude $1.0 \mathrm{V} ?$

Short Answer

Expert verified
Answer: The coil must be rotated with an angular velocity of approximately 1273 s⁻¹ to generate an emf of amplitude 1.0 V at the given location.

Step by step solution

01

Convert the units to MKS system

Convert the radius from centimeters to meters and Earth's magnetic field from millitesla (mT) to tesla (T). - Radius: \(5.0\,cm = 0.05\,m\) - Magnetic field: \(0.50\,mT = 0.0005\,T\)
02

Write Faraday's Law of electromagnetic induction

Faraday's Law states that the induced emf in a coil is equal to the rate of change of magnetic flux through the coil. The formula for the emf induced in a coil with N turns is: \(\epsilon = N \frac{d\Phi_{B}}{dt}\)
03

Calculate the magnetic flux through the coil

The magnetic flux through a circular loop of area A and magnetic field B in the direction perpendicular to the loop is given by: \(\Phi_{B} = B A\) Here, the area of the circular loop can be calculated as \(A = \pi r^2\).
04

Differentiate the magnetic flux with respect to time

By differentiating the magnetic flux with respect to time, we will obtain the rate of change of magnetic flux through the coil: \(\frac{d\Phi_{B}}{dt} = \frac{d(B A )}{dt}\) Keeping the value of B constant, the derivative of the area of the coil with respect to time becomes the variable part. As the coil is rotating, the area changes with respect to time, so we should find the area as a function of time, i.e., \(A(t) = A\sin(\omega t)\). Differentiating this with respect to time, we get: \(\frac{dA(t)}{dt} = A\omega\cos(\omega t)\)
05

Calculate angular velocity

To find the required angular velocity, plug the values of N, B, A, and the amplitude of the emf (1.0 V) back into the Faraday's Law equation: \(1.0\,V = 1000\,turns \times 0.0005\,T \times (\pi (0.05\,m)^2)\omega\cos(\omega t)\) For the peak emf, we have \(\cos(\omega t) = 1\). After simplifying the equation, we can solve for the angular velocity \(\omega\): \(\omega = \frac{1.0\,V}{1000\,turns \times 0.0005\,T \times (\pi (0.05\,m)^2)}\) After calculations, we get:
06

Final Answer

\(\omega \approx 1273\,\mathrm{s}^{-1}\) Hence, the coil must be rotated with an angular velocity of approximately 1273 s⁻¹ to generate an emf of amplitude 1.0 V at the given location.

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