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Chapter 14: Problem 2015

The magnetic flux linked with a coil, in webers, is given by the equation \(\Phi=4 t^{2}-t+7\). Then the magnitude of induced emf at 2 sec will be ... (a) \(15 \mathrm{v}\) (b) \(19 \mathrm{v}\) (c) \(17 \mathrm{v}\) (d) \(21 \mathrm{v}\)

Short Answer

Expert verified
The magnitude of the induced emf at 2 seconds is \(15 V\).

Step by step solution

01

Differentiate the given magnetic flux equation with respect to time

To find the rate of change of magnetic flux, differentiate the given equation with respect to time (t): Given equation: \(\Phi=4t^2 - t + 7\) Now, differentiate it with respect to 't': \(\frac{d\Phi}{dt} = \frac{d}{dt}(4t^2 - t + 7)\) Applying the differentiation, we get: \(\frac{d\Phi}{dt} = 8t - 1\)
02

Evaluate the expression at t = 2 sec

Substitute t = 2 sec in the equation from Step 1: \(\frac{d\Phi}{dt} = 8(2) - 1\) Calculating the value, we get: \(\frac{d\Phi}{dt} = 16 - 1\) \(\frac{d\Phi}{dt} = 15\)
03

Apply Faraday's Law to find the induced emf

According to Faraday's Law, the induced emf (E) is given by: \(E = -\frac{d\Phi}{dt}\) We found that \(\frac{d\Phi}{dt} = 15\) in Step 2, so the induced emf (E) can be calculated as follows: \(E = -15\)
04

Find the magnitude of the induced emf

The magnitude of induced emf is the absolute value of E obtained in Step 3: \(Magnitude\: of\: E = |-15|\) Calculating the magnitude, we get: \(Magnitude\: of\: E = 15 V\) Therefore, at 2 seconds, the magnitude of the induced emf is 15 V. The correct answer is (a) \(15 V\).

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