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Chapter 14: Problem 2017

A coil has an area of \(0.05 \mathrm{~m}^{2}\) \& it has 800 turns. It is placed perpendicularly in a magnetic field of strength, $4 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{-2}\(. It is rotated through \)90^{\circ}\( in \)0.1 \mathrm{sec}$. The average emf induced in the coil is... (a) \(0.056 \mathrm{v}\) (b) \(0.046 \mathrm{v}\) (c) \(0.026 \mathrm{v}\) (d) \(0.016 \mathrm{v}\)

Short Answer

Expert verified
The short answer is: The average emf induced in the coil is \(0.016 \mathrm{V}\).

Step by step solution

01

Calculate the initial magnetic flux

We first need to find the initial magnetic flux in the coil when it is placed perpendicularly to the magnetic field. The formula for magnetic flux is: $$ \phi = BA\cos\theta $$ Where: - \(\phi\) is the magnetic flux - \(B\) is the magnetic field strength - \(A\) is the area of the coil - \(\theta\) is the angle between the magnetic field and the area vector of the coil Initially, when the coil is placed perpendicularly to the magnetic field (\(\theta = 0^\circ\)), we have: $$ \phi_0 = (4 \times 10^{-5} \mathrm{Wb/m^2})(0.05 \mathrm{m^2})\cos(0) = 2\times 10^{-6}\mathrm{Wb} $$
02

Calculate the final magnetic flux

Next, we need to find the magnetic flux when the coil is rotated through 90 degrees. When the coil is rotated (\(\theta = 90^\circ\)), we have: $$ \phi_f = (4 \times 10^{-5} \mathrm{Wb/m^2})(0.05 \mathrm{m^2})\cos(90) = 0\mathrm{Wb} $$
03

Calculate the change in magnetic flux

Now that we have the initial and final magnetic flux, we can calculate the change in magnetic flux, which is also known as the magnetic flux through the coil: $$ \Delta\phi = \phi_f - \phi_0 = 0 - 2\times 10^{-6}\mathrm{Wb} = -2\times 10^{-6}\mathrm{Wb} $$
04

Calculate the induced emf

Finally, we can use Faraday's law to calculate the induced emf: $$ \varepsilon = -N \frac{d\phi}{dt} = -800 \frac{-2\times 10^{-6}\mathrm{Wb}}{0.1 \mathrm{s}} = 0.016 \mathrm{V} $$ Therefore, the answer is (d) \(0.016 \mathrm{V}\).

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Most popular questions from this chapter

The value of alternating emf \(E\) in the given ckt will be. (a) \(100 \mathrm{~V}\) (b) \(20 \mathrm{~V}\) (c) \(220 \mathrm{~V}\) (d) \(140 \mathrm{~V}\)

A coil having n turns \(\&\) resistance \(R \Omega\) is connected with a galvanometer of resistance \(4 \mathrm{R} \Omega\). This combination is moved from a magnetic field \(\mathrm{W}_{1} \mathrm{~Wb}\) to $\mathrm{W}_{2} \mathrm{~Wb}\( in \)\mathrm{t}$ second. The induced current in the circuit is.... (a) $-\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{5 \mathrm{Rnt}\\}\right]$ (b) $-\mathrm{n}\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{5 \mathrm{Rt}\\}\right]$ (c) $-\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{\mathrm{Rnt}\\}\right]$ (d) $-\mathrm{n}\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{\mathrm{Rt}\\}\right]$

A 20 volts ac is applied to a circuit consisting of a resistance and a coil with negligible resistance. If the voltage across the resistance is $12 \mathrm{~V}$, the voltage across the coil is, (a) 16 volts (b) 10 volts (c) 8 volts (d) 6 volts

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