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Chapter 29: Problem 42

A short coil with radius \(R=10.0 \mathrm{~cm}\) contains \(N=30.0\) turns and surrounds a long solenoid with radius \(r=8.00 \mathrm{~cm}\) containing \(n=60\) turns per centimeter. The current in the short coil is increased at a constant rate from zero to \(i=2.00 \mathrm{~A}\) in a time of \(t=12.0 \mathrm{~s}\). Calculate the induced potential difference in the long solenoid while the current is increasing in the short coil.

Short Answer

Expert verified
Question: Calculate the induced potential difference in a long solenoid due to the increasing current in a surrounding short coil from 0 to i in a time t. Answer: The induced potential difference in the long solenoid can be calculated using Faraday's Law of Electromagnetic Induction: \(\text{Potential Difference} = -\frac{d\Phi_{\text{total}}}{dt}\) Where \(\Phi_{\text{total}}\) is the total magnetic flux through the solenoid due to the current in the short coil and can be found using the steps provided in the solution.

Step by step solution

01

Calculate the magnetic field inside the solenoid due to the current in the short coil

First, we will calculate the magnetic field inside the solenoid due to the current in the short coil using the Biot-Savart Law. The magnetic field inside a solenoid is given by: \(B = \frac{\mu_0 ni}{2}\) Where \(\mu_0\) is the permeability of free space \((4\pi \times 10^{-7} T\text{m}/\text{A})\), \(n\) is the number of turns per centimeter in the solenoid, and \(i\) is the current in the short coil.
02

Calculate the magnetic flux through a single turn of the solenoid

Next, we will calculate the magnetic flux through a single turn of the solenoid. The magnetic flux is given by: \(\Phi = BA\) Where \(A\) is the area of the solenoid's cross-section, which can be calculated using the formula: \(A = \pi r^2\)
03

Calculate the total magnetic flux through the solenoid

Now, we will calculate the total magnetic flux through the solenoid by multiplying the magnetic flux through a single turn by the total number of turns N in the short coil: \(\Phi_{\text{total}} = N\Phi\)
04

Calculate the rate of change of magnetic flux

To calculate the induced potential difference, we need to find the rate of change of the magnetic flux. As the current in the short coil is increasing at a constant rate with time, we can calculate the rate of change of magnetic flux as follows: \(\frac{d\Phi_{\text{total}}}{dt} = N\frac{d\Phi}{dt}\)
05

Calculate the induced potential difference in the long solenoid

Now, we can use Faraday's Law of Electromagnetic Induction to calculate the induced potential difference in the long solenoid: \(\text{Potential Difference} = -\frac{d\Phi_{\text{total}}}{dt}\) Substitute the values from steps 1 to 4 into the above equation to find the induced potential difference in the long solenoid.

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Most popular questions from this chapter

An 8 -turn coil has square loops measuring \(0.200 \mathrm{~m}\) along a side and a resistance of \(3.00 \Omega\). It is placed in a magnetic field that makes an angle of \(40.0^{\circ}\) with the plane of each loop. The magnitude of this field varies with time according to \(B=1.50 t^{3}\), where \(t\) is measured in seconds and \(B\) in teslas. What is the induced current in the coil at \(t=2.00 \mathrm{~s} ?\)

A respiration monitor has a flexible loop of copper wire, which wraps about the chest. As the wearer breathes, the radius of the loop of wire increases and decreases. When a person in the Earth's magnetic field (assume \(\left.0.426 \cdot 10^{-4} \mathrm{~T}\right)\) inhales, what is the average current in the loop, assuming that it has a resistance of \(30.0 \Omega\) and increases in radius from \(20.0 \mathrm{~cm}\) to \(25.0 \mathrm{~cm}\) over \(1.00 \mathrm{~s}\) ? Assume that the magnetic field is perpendicular to the plane of the loop.

Which of the following statements regarding self induction is correct? a) Self-induction occurs only when a direct current is flowing through a circuit. b) Self-induction occurs only when an alternating current is flowing through a circuit. c) Self-induction occurs when either a direct current or an alternating current is flowing through a circuit. d) Self-induction occurs when either a direct current or an alternating current is flowing through a circuit as long as the current is varying.

A rectangular wire loop (dimensions of \(h=15.0 \mathrm{~cm}\) and \(w=8.00 \mathrm{~cm}\) ) with resistance \(R=5.00 \Omega\) is mounted on a door. The Earth's magnetic field, \(B_{\mathrm{E}}=2.6 \cdot 10^{-5} \mathrm{~T}\), is uniform and perpendicular to the surface of the closed door (the surface is in the \(x z\) -plane). At time \(t=0,\) the door is opened (right edge moves toward the \(y\) -axis) at a constant rate, with an opening angle of \(\theta(t)=\omega t,\) where \(\omega=3.5 \mathrm{rad} / \mathrm{s}\) Calculate the direction and the magnitude of the current induced in the loop, \(i(t=0.200 \mathrm{~s})\).

A rectangular conducting loop with dimensions \(a\) and \(b\) and resistance \(R\), is placed in the \(x y\) -plane. A magnetic field of magnitude \(B\) passes through the loop. The magnetic field is in the positive \(z\) -direction and varies in time according to \(B=B_{0}\left(1+c_{1} t^{3}\right),\) where \(c_{1}\) is a constant with units of \(1 / \mathrm{s}^{3}\) What is the direction of the current induced in the loop, and what is its value at \(t=1 \mathrm{~s}\) (in terms of \(a, b, R, B_{0},\) and \(\left.c_{1}\right) ?\)
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