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Chapter 34: Problem 59

A 2.0-cm-diameter solenoid is wrapped with 1000 turns per meter. \(0.50 \mathrm{cm}\) from the axis, the strength of an induced electric field is \(5.0 \times 10^{-4} \mathrm{V} / \mathrm{m} .\) What is the rate \(d I / d t\) with which the current through the solenoid is changing?

Short Answer

Expert verified
The rate of change of current through the solenoid is \(dI/dt = -200 \, A/s\). This means the current is decreasing at a rate of 200 Ampere per second.

Step by step solution

01

Identify given parameters and target value

In this problem, we are given the radius of the solenoid \(R = 1.0 \, cm = 0.01 \, m\). We are also given the number of turns per unit length of the solenoid \(n = 1000 \, turns/m\), the distance from the axis, \(r = 0.50 \, cm = 0.005 \, m\), and the induced electric field strength at that distance \(E = 5.0 \times 10^{-4} \, V/m\). We are asked to find the rate of change of current through the solenoid, \(dI/dt\).
02

Apply Ampere's Law

First, we know that the magnetic field \(B = \mu_0 n I\), where \(\mu_0 = 4\pi \times 10^{-7} \, T*m/A\) is the permeability of free space. Ampere's law states that the line integral of the electric field around a closed path is equal to the rate of change of the magnetic flux through the area enclosed by the path. In mathematical terms, \(E \cdot 2\pi r = - d(B \cdot A)/dt\). Here, \(B = \mu_0 n I\) is the magnetic field inside the solenoid, and \(A = \pi r^{2}\) is the cross-sectional area of the region where \(E\) is measured.
03

Solve for target value

Substituting the magnetic field expression into Amp猫re's law, we have \(E = - (d/dt)(\mu_0 n I) * (\pi r^{2}) / (2\pi r)\), which simplifies to \(E = - (r/2) d(\mu_0 n I) / dt\). Solving for \(dI/dt\), we have \(dI/dt = - 2E /(r \mu_0 n)\). Substituting the given values, we find \(dI/dt = - (2 * 5.0 \times 10^{-4} \, V/m) / (0.005 \, m * 4 \pi \times 10^{-7} T*m/A * 1000 \, turns/m) = -200 \, A/s\). Note that the minus sign indicates the current is decreasing.

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Most popular questions from this chapter

A 50-turn, 4.0-cm-diameter coil with \(R=0.50 \Omega\) surrounds a 2.0-cm- diameter salenoid. The solenoid is \(20 \mathrm{cm}\) long and has 200 turns. The \(60 \mathrm{Hz}\) current through the solenoid is \(I_{\mathrm{eol}}=\) \((0.50 \mathrm{A}) \sin (2 \pi f t) .\) Find an expression for \(I_{\text {coil }},\) the induced current in the coil as a function of time.

A 10-turn coil of wire having a diameter of \(1.0 \mathrm{cm}\) and a resistance of \(0.20 \Omega\) is in a \(1.0 \mathrm{mT}\) magnetic field, with the coil oriented for maximum flux. The coil is connected to an uncharged \(1.0 \mu \mathrm{F}\) capacitor rather than to a current meter. The coil is quickly pulled out of the magnetic field. Afterward, what is the voltage across the capacitor? Hint: Use \(I=d q / d t\) to relate the net change of flux to the amount of charge that flows to the capacitor.

A 10-cm-long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire and rail is \(0.20 \Omega .\) Pulling the wire with a force of \(1.0 \mathrm{N}\) causes 4.0 W of power to be dissipated in the circuit. a. What is the speed of the wire when pulled with \(1.0 \mathrm{N} ?\) b. What is the strength of the magnetic field?

An \(L C\) circuit is built with a \(20 \mathrm{mH}\) inductor and an \(8.0 \mu \mathrm{F}\) capacitor. The current has its maximum value of 0.50 A at \(t=0\) s. a. How long is it until the capacitor is fully charged? b. What is the voltage across the capacitor at that time?

The earth's magnetic field strength is \(5.0 \times 10^{-5}\) T. How fast would you have to drive your car to create a \(1.0 \mathrm{V}\) motional \(\mathrm{cmf}\) along your 1.0-m-long radio antenna? Assume that the motion of the antenna is perpendicular to \(\vec{B}\).
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