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Chapter 16: Problem 17

When the current changes from \(+2 \mathrm{~A}\) to \(-2 \mathrm{~A}\) in \(0.05 \mathrm{~s}\), an EMF of \(8 \mathrm{~V}\) is induced in a coil. The coefficient of self-induction of the coil is (A) \(0.1 \mathrm{H}\) (B) \(0.2 \mathrm{H}\) (C) \(0.4 \mathrm{H}\) (D) \(0.8 \mathrm{H}\)

Short Answer

Expert verified
The coefficient of self-induction of the coil is (A) \(0.1 \mathrm{H}\).

Step by step solution

01

Write down the given values

From the problem, we are given: 1. Initial current, \(I_1 = +2A\) 2. Final current, \(I_2 = -2A\) 3. Time taken for the change, \(\Delta t = 0.05s\) 4. EMF induced, \(E = 8V\) Our goal is to find the coefficient of self-induction, \(L\), of the coil.
02

Calculate the change in current

In order to find the coefficient of self-induction, we need to first find the change in current. The change in current, \(\Delta I\), can be calculated using the initial and final currents: \(\Delta I = I_2 - I_1\)
03

Calculate the rate of change of current

Now, we need to find the rate of change of current, \(\frac{\Delta I}{\Delta t}\). This can be calculated using the change in current and the time taken for the change: \(\frac{\Delta I}{\Delta t} = \frac{I_2 - I_1}{\Delta t}\)
04

Apply Faraday's Law of Electromagnetic Induction

According to Faraday's Law of Electromagnetic Induction, the induced EMF in a coil is directly proportional to the rate of change of magnetic flux through the coil. In the case of self-induction, the EMF can be described using the coefficient of self-induction and the rate of change of current: \(E = L \cdot \frac{\Delta I}{\Delta t}\)
05

Solve for the coefficient of self-induction

Now, we can plug in the given values and solve for the coefficient of self-induction, \(L\): \(L = \frac{E}{\frac{\Delta I}{\Delta t}}\)
06

Calculate the values and select the answer

First, we need to find \(\Delta I\) and \(\frac{\Delta I}{\Delta t}\): \(\Delta I = I_2 - I_1 = -2A - 2A = -4A\) \(\frac{\Delta I}{\Delta t} = \frac{-4A}{0.05s} = -80A/s\) Now, we can plug in these values and the given EMF to find the coefficient of self-induction, \(L\): \(L = \frac{8V}{-80A/s} = -0.1H\) Since the coefficient of self-induction cannot be negative, we'll take the absolute value: \(L = 0.1H\) Thus, the correct answer is (A) \(0.1 \mathrm{H}\).

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Most popular questions from this chapter

A coil of \(20 \times 20 \mathrm{~cm}\) having 30 turns is making \(30 \mathrm{rps}\) in a magnetic field of 1 tesla. The peak value of the induced EMF is approximately (A) \(452 \mathrm{~V}\) (B) \(226 \mathrm{~V}\) (C) \(113 \mathrm{~V}\) (D) \(339 \mathrm{~V}\)

A magnetic field \(\vec{B}=\left(\frac{B_{0} y}{a}\right) \hat{k}\) is into the paper in the \(+z\) direction. \(B_{0}\) and \(a\) are positive constants. A square loop \(E F G H\) of side \(a\), mass \(m\), and resistance \(R\), in \(x-y\) plane, starts falling under the influence of gravity. Assume \(x\)-axis is horizontal and \(y\) is vertically downward. The magnitude and direction of the net Lorentz force, acting on the loop when its speed is \(v\), is (A) \(\frac{B_{0} a^{2} v}{R}\), upward (B) \(\frac{B_{0} a^{2} v}{R}\), downward (C) \(\frac{B_{0}^{2} a^{2} v}{R}\), downward (D) \(\frac{B_{0}^{2} a^{2} v}{R}\), upward

Two particles each of mass \(m\) and charge \(q\) are attached to the two ends of a light rigid rod of length \(2 \ell\). The rod is rotated at a constant angular speed about a perpendicular axis passing through its centre. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod is (A) \(\frac{q}{2 m} \quad\) (B) \(\frac{q}{m}\) (C) \(\frac{2 q}{m}\) (D) \(\frac{q}{\pi m}\)

Two straight long conductors \(A O B\) and \(C O D\) are perpendicular to each other and carry currents \(I_{1}\) and \(I_{2}\), respectively. The magnitude of the magnetic induction at a point \(P\) at a distance \(a\) from the point \(O\) in a direction perpendicular to the plane \(A B C D\) is (A) \(\frac{\mu_{0}}{2 \pi a}\left(I_{1}+I_{2}\right)\) (B) \(\frac{\mu_{0}}{2 \pi a}\left(I_{1}-I_{2}\right)\) (C) \(\frac{\mu_{0}}{2 \pi a}\left(I_{1}^{2}+I_{2}^{2}\right)^{1 / 2}\) (D) \(\frac{\mu_{0}}{2 \pi a}\left(\frac{I_{1} I_{2}}{I_{1}+I_{2}}\right)\)

Two parallel vertical metallic rails \(A B\) and \(C D\) are separated by \(1 \mathrm{~m}\). They are connected at the two ends by resistances \(R_{1}\) and \(R_{2}\) as shown. A horizontal metallic bar \(P Q\) of mass \(0.2 \mathrm{~kg}\) slides without friction, vertically down the rails under the action of gravity. There is uniform horizontal magnetic field of \(0.6 \mathrm{~T}\) perpendicular to plane of the rails. It is observed that 12 when the terminal velocity attained, the power dissipated in \(R_{1}\) and \(R_{2}\) are \(0.76 \mathrm{~W}\) and \(1.2 \mathrm{~W}\), respectively. Find the terminal velocity of bar in \(\mathrm{m} / \mathrm{s}\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)\).
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