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Chapter 16: Problem 48

If the flux of magnetic induction through a coil of resistance \(R\) and having \(n\) turns changes from \(\phi_{1}\) to \(\phi_{2}\), then the magnitude of the charge that passes through the coil is (A) \(\frac{\left(\phi_{2}-\phi_{1}\right)}{R}\) (B) \(\frac{n\left(\phi_{2}-\phi_{1}\right)}{R}\) (C) \(\frac{\left(\phi_{2}-\phi_{1}\right)}{n R}\) (D) \(\frac{n R}{\left(\phi_{2}-\phi_{1}\right)}\)

Short Answer

Expert verified
The short answer is: The magnitude of the charge that passes through the coil is given by (B): \(\frac{n(\phi_{2} - \phi_{1})}{R}\).

Step by step solution

01

Write down Faraday's Law of Electromagnetic Induction

Faraday's Law of Electromagnetic Induction states that the induced electromotive force (E) in a coil is equal to the negative of the rate of change of the magnetic flux (桅): \[ E = -n\frac{d\Phi}{dt} \]
02

Calculate the induced electromotive force (EMF)

As we want to find the magnitude of the change that passes through the coil, we will not consider the negative sign in the formula. In this case, we have: \[ E = n\frac{d\Phi}{dt} = n\frac{\phi_{2} - \phi_{1}}{\Delta t} \]
03

Calculate the current passing through the coil

We know that the current, I, passing through a coil with resistance R and EMF, E, is given by Ohm's Law: \[ I = \frac{E}{R} \] Substituting the value of E obtained in Step 2, we get: \[ I = \frac{n(\phi_{2} - \phi_{1})}{R\Delta t} \]
04

Calculate the magnitude of the charge through the coil

The magnitude of the charge, Q, passing through the coil is given by the product of the current (I) and the time interval (螖t): \[ Q = I \Delta t \] Substituting the value of I obtained in Step 3, we get: \[ Q = \frac{n(\phi_{2} - \phi_{1})}{R\Delta t} \times \Delta t = \frac{n(\phi_{2} - \phi_{1})}{R} \] Thus, the magnitude of the charge that passes through the coil is given by (B): \[\frac{n(\phi_{2} - \phi_{1})}{R}\]

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Most popular questions from this chapter

A metal conductor of length \(1 \mathrm{~m}\) rotates vertically about one of its ends at angular velocity 5 radians per second. If the horizontal component of earth's magnetic field is \(0.2 \times 10^{-4} \mathrm{~T}\), then the EMF developed between the two ends of the conductor is (A) \(5 \mathrm{mV}\) (B) \(50 \mu \mathrm{V}\) (C) \(5 \mu \mathrm{V}\) (D) \(50 \mathrm{mV}\)

The core of any transformer is laminated so as to (A) reduce the energy loss due to eddy currents. (B) make it light weight. (C) make it robust and strong. (D) increase the secondary voltage.

In a coil when current changes from \(10 \mathrm{~A}\) to \(2 \mathrm{~A}\) in time \(0.1 \mathrm{~s}\), induced EMF is \(3.20 \mathrm{~V}\). The self-inductance of coil is (A) \(4 \mathrm{H}\) (B) \(0.4 \mathrm{H}\) (C) \(0.04 \mathrm{H}\) (D) \(5 \mathrm{H}\)

A coil of inductance \(1 H\) and resistance \(10 \Omega\) is connected to a resistance-less battery of EMF \(50 \mathrm{~V}\) at time \(t=0 .\) The ratio of rate at which magnetic energy is stored in the coil to the rate at which energy is supplied by the battery at \(t=0.1 \mathrm{~s} .\) is \(x \times 10^{-2}\). Find the value of \(x\). (Given \(\left.\frac{1}{e}=0.37\right)\)

Net force on a current carrying loop kept in uniform magnetic field is zero and the torque on the loop \(\vec{\tau}=\vec{M} \times \vec{B}\), where \(M\) and \(B\) are magnetic dipole moment and magnetic field intensity, respectively. If it is free to rotate, then it will rotates about an axis passing through its centre of mass and parallel to \(\vec{\tau}\). Potential energy of the loop is given by \(U=-\vec{M} \cdot \vec{B}\). Assume a current carrying ring with its centre at the origin and having moment of inertia \(2 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^{2}\) about an axis passing through one of its diameter and magnetic moment \(\vec{M}=(3 \hat{i}-4 \hat{j}) \mathrm{Am}^{2}\). At time \(t=0\), a magnetic field \(\vec{B}=(4 \hat{i}-3 \hat{j}) T\) is switched on. Then Maximum angular velocity of the ring (in \(\mathrm{rad} / \mathrm{s})\) will be (A) \(50 \sqrt{2}\) (B) \(25 \sqrt{2}\) (C) \(100 \sqrt{2}\) (D) \(150 \sqrt{2}\)
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