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Chapter 16: Problem 44

A varying magnetic flux linking a coil is given by \(\phi=x t^{2}\). If at a time \(t=3 \mathrm{~s}\), the EMF induced is \(9 \mathrm{~V}\), then the value of \(x\) is (A) \(0.66 \mathrm{Wbs}^{-2}\) (B) \(1.5 \mathrm{Wbs}^{-2}\) (C) \(-0.66 \mathrm{Wbs}^{-2}\) (D) \(-1.5 \mathrm{Wbs}^{-2}\)

Short Answer

Expert verified
The value of x in the given magnetic flux equation is \( -1.5 \ \mathrm{Wbs}^{-2}\). Therefore, the correct answer is (D) \(-1.5 \mathrm{Wbs}^{-2}\).

Step by step solution

01

Write down the given information

We have the magnetic flux given by the equation: \(\phi = xt^2\), EMF induced at time t = 3 seconds: E = 9 V. Our task is to find the value of x from the given information.
02

Find the equation of induced EMF

According to Faraday's law of electromagnetic induction, the induced EMF is equal to the negative rate of change of magnetic flux with respect to time. Mathematically, \[-E = k \frac{d\phi}{dt},\] where k is the constant of proportionality, and in this case, k = 1 because one loop of wire is assumed. Therefore, the equation becomes \[-E = \frac{d\phi}{dt}.\]
03

Calculate the derivative of the magnetic flux

Now we need to find the derivative of the given magnetic flux equation with respect to time. Given \[\phi = xt^2,\] Differentiate both sides with respect to time t: \[\frac{d\phi}{dt} = x\frac{d(t^2)}{dt},\] \[\frac{d\phi}{dt} = 2xt.\]
04

Substitute the given values and solve for x

Now, substitute the given values of induced EMF E and time t in the equation and solve for x. From Step 2, we have \[-E = \frac{d\phi}{dt}.\] Substitute the values, \[-9 = 2x(3),\] \[-9 = 6x,\] \[x = -\frac{9}{6},\] \[x = -1.5 \ \mathrm{Wbs}^{-2}.\]
05

Answer the question

The value of x in the given magnetic flux equation is \( -1.5 \ \mathrm{Wbs}^{-2}.\) Therefore, the correct answer is (D) \(-1.5 \mathrm{Wbs}^{-2}\).

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