A circular coil is rotating in a magnetic field of magnitude 0.25T with angular speed 6 rpm about its diameter. At \( t = 0 \), the coil’s configuration is as shown. If the induced emf after the coil is rotated by an angle of 30° is 1.6 mV, find the radius of the coil (in cm). (\( \pi^2 = 10 \))
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Use the formula for induced emf and remember to convert angular speed to radians per second. The induced emf depends on the area of the coil, magnetic field strength, and the rate of rotation.
To find the radius of the coil, we start with Faraday's law of electromagnetic induction, which states that the induced emf \( \varepsilon \) in a coil is given by the rate of change of magnetic flux \( \Phi \) through the coil:
\(\varepsilon = -\frac{d\Phi}{dt}\)
The magnetic flux \( \Phi \) is given by:
\(\Phi = B \cdot A \cdot \cos(\theta)\)
where \( B = 0.25 \, \text{T} \) is the magnetic field strength, \( A \) is the area of the coil, and \( \theta \) is the angle between the magnetic field and the normal to the coil's plane.
The area \( A \) of the circular coil is:
\( A = \pi r^2 \)
Given that the coil rotates with an angular speed of \( \omega = 6 \, \text{rpm} = \frac{6 \times 2\pi}{60} \, \text{rad/s} = 0.2 \pi \, \text{rad/s}\), the angle \( \theta \) changes with time as \( \theta = \omega t\). After rotation by 30° (\(\theta = 30^\circ = \frac{\pi}{6}\)), and \(\varepsilon = 1.6 \, \text{mV} = 1.6 \times 10^{-3} \, \text{V}\).
Now, find \( \varepsilon \) using the induced emf formula:
\(\varepsilon = \left| \frac{d}{dt}(B \pi r^2 \cos(\omega t)) \right|\)
\(\varepsilon = \left| -B \pi r^2 \omega \sin(\omega t) \right|\)
At \( t \) when \(\theta = \frac{\pi}{6}\),
\(\sin(\omega t) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\)
Substitute the values, we get:
\(1.6 \times 10^{-3} = 0.25 \times \pi \times r^2 \times 0.2\pi \times \frac{1}{2}\)
Solve for \( r^2 \):
\(1.6 \times 10^{-3} = 0.25 \times \pi \times r^2 \times 0.1\pi\)
\(r^2 = \frac{1.6 \times 10^{-3}}{0.025 \times \pi^2}\)
Using \(\pi^2 = 10\) from given:
\(r^2 = \frac{1.6 \times 10^{-3}}{0.025 \times 10}\)
\(r^2 = \frac{1.6 \times 10^{-3}}{0.25}\)
\(r^2 = 6.4 \times 10^{-3}\)
\(r = \sqrt{6.4 \times 10^{-3}} \approx 0.08 \, \text{m} = 8 \, \text{cm}\)
Thus, the radius of the coil is \(8 \, \text{cm}\), fitting the range 8,8.
