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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

People with pacemakers or other mechanical devices as implants are often warned to stay away from large machinery or motors. Why?

Consider a clinical MRI (magnetic resonance imaging) superconducting mag. net has a diameter of \(1.00 \mathrm{~m}\) length of \(1.50 \mathrm{~m}\), and a uniform magnetic field of 3.00 T. Determine (a) the energy density of the magnetic field and (b) the total energy in the solenoid.

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 \(\left.c_{1}\right) ?\)

Faraday's Law of Induction states a) that a potential difference is induced in a loop when there is a change in the magnetic flux through the loop. b) that the current induced in a loop by a changing magnetic field produces a magnetic field that opposes this change in magnetic field. c) that a changing magnetic field induces an electric field. d) that the inductance of a device is a measure of its opposition to changes in current flowing through it. e) that magnetic flux is the product of the average magnetic field and the area perpendicular to it that it penetrates.

At Los Alamos National Laboratories, one means of producing very large magnetic fields is known as the EPFCG (explosively-pumped flux compression generator), which is used to study the effects of a high-power electromagnetic pulse (EMP) in electronic warfare. Explosives are packed and detonated in the space between a solenoid and a small copper cylinder coaxial with and inside the solenoid, as shown in the figure. The explosion occurs in a very short time and collapses the cylinder rapidly. This rapid change creates inductive currents that keep the magnetic flux constant while the cylinder's radius shrinks by a factor of \(r_{\mathrm{i}} / r_{\mathrm{f}}\). Estimate the magnetic field produced, assuming that the radius is compressed by a factor of 14 and the initial magnitude of the magnetic field, \(B_{i}\), is \(1.0 \mathrm{~T}\).
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