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Chapter 16: Problem 49

A varying magnetic flux linking a coil is given by \(\phi=3 t^{2}\). The magnitude of induced EMF in the loop at \(t=3 \mathrm{~s}\) is (A) \(3 \mathrm{~V}\) (B) \(9 \mathrm{~V}\) (C) \(18 \mathrm{~V}\) (D) \(27 \mathrm{~V}\)

Short Answer

Expert verified
The induced EMF magnitude at \(t=3s\) is found using Faraday's law and the given magnetic flux function \(\phi=3t^2\). After finding the derivative with respect to time, \(\frac{d\phi}{dt} = 6t\), we substitute the given time value and get the induced EMF magnitude: \(|\varepsilon| = 18V\). Thus, the correct answer is (C) \(18 \mathrm{~V}\).

Step by step solution

01

Find the derivative of the magnetic flux function

Faraday's law states that the induced EMF is equal to the negative rate of change of magnetic flux with respect to time, i.e., \[ \varepsilon = -\frac{d\phi}{dt} \] Given the magnetic flux function: \[ \phi = 3t^2 \] We need to find the derivative of this function with respect to time \(t\). Using the power rule for differentiation, we get: \[ \frac{d\phi}{dt} = \frac{d(3t^2)}{dt} = 6t \]
02

Find the induced EMF at t = 3 seconds

Now that we have the derivative of the magnetic flux function, we need to find the induced EMF at \(t = 3\) seconds. Substitute the given time value into the derivative we found in Step 1: \[ \frac{d\phi}{dt} = 6(3) = 18 \] Since the induced EMF is equal to the negative rate of change of the magnetic flux, we have: \[ \varepsilon = -\frac{d\phi}{dt} = -18 \] Since we are asked for the magnitude of the induced EMF, we can disregard the negative sign and report the value as: \[ |\varepsilon| = 18 V \]
03

Identify the correct option

From our calculations, the magnitude of induced EMF at \(t = 3\) seconds is \(18V\). Comparing our result with the given options, we find that option (C) matches our calculated value: (C) \(18 \mathrm{~V}\) Therefore, the correct answer is (C) \(18 \mathrm{~V}\).

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Most popular questions from this chapter

Figure \(16.66\) shows four rods having \(\lambda=0.5 \Omega / \mathrm{m}\) resistance per unit length. The arrangement is kept in a magnetic field of constant magnitude \(B=2 \mathrm{~T}\) and directed perpendicular to the plane of the figure and directed inwards. Initially, the rods form a square of side length \(\ell=15 \mathrm{~m}\) as shown. Now each wire starts moving with constant velocity \(v=5 \mathrm{~m} / \mathrm{s}\) towards opposite wire. Find the force required in newton on each wire to keep its velocity constant at \(t=1 \mathrm{~s}\).

The EMF induced in a 1 millihenry inductor in which the current changes from \(5 \mathrm{~A}\) to \(3 \mathrm{~A}\) in \(10^{-3}\) second is (A) \(2 \times 10^{-6} \mathrm{~V}\) (B) \(8 \times 10^{-6} \mathrm{~V}\) (C) \(2 \mathrm{~V}\) (D) \(8 \mathrm{~V}\)

A capacitance \(C\) is connected to a conducting rod of length \(\ell\) moving with a velocity \(v\) in a transverse magnetic field \(B\) then the charge developed in the capacitor is (A) Zero (B) \(B \ell v C\) (C) \(\frac{B l v C}{2}\) (D) \(\frac{B l v C}{3}\)

The magnetic susceptibility of a material of a rod is 499. Permeability of vacuum is \(4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\). Absolute permeability of the material of the rod in henry per meter is (A) \(\pi \times 10^{-4}\) (B) \(2 \pi \times 10^{-4}\) (C) \(3 \pi \times 10^{-4}\) (D) \(4 \pi \times 10^{-4}\)

In a LCR circuit, capacitance is changed from \(\mathrm{C}\) to \(2 \mathrm{C}\), For the resonant frequency to remain unchanged, the inductance should be changed from \(L\) to (A) \(\mathrm{L} / 2\) (B) \(2 \mathrm{~L}\) (C) \(4 \mathrm{~L}\) (D) \(\underline{L} / 4\)
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