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Chapter 27: Problem 39

A 2000 -turn solenoid is \(2.0 \mathrm{m}\) long and \(15 \mathrm{cm}\) in diameter. The solenoid current is increasing at \(1.0 \mathrm{kA} / \mathrm{s} .\) (a) Find the current in a 10 -cm-diameter wire loop with resistance \(5.0 \Omega\) lying inside the solenoid and perpendicular to the solenoid axis. (b) Repeat for a similarly oriented \(25-\mathrm{cm}\) -diameter loop with the same resistance, lying entirely outside the solenoid.

Short Answer

Expert verified
The currents in the \(10 cm\) diameter loop inside the solenoid and the \(25 cm\) diameter loop outside the solenoid are \(0.628 A\) and \(1.77 A\) respectively.

Step by step solution

01

Calculate the magnetic field inside the solenoid

Firstly, the magnitude of the magnetic field inside the solenoid is necessary. As the solenoid is long, it can be assumed that the field inside is constant and parallel to the length of the solenoid. The formula for the calculation of this field is given by \(B=\mu_0*n*I\), where \(B\) is the magnetic field, \(\mu_0\) is the permeability of free space, \(n\) is the number of turns per unit length of the solenoid, and \(I\) is the current. In this case, \(n=2000/2\) turns/meter, and \(\mu_0 = 4*\pi *10^{-7} T*m/A\). However, as current is changing, \(B\) will be \(B=\mu_0 * n * di/dt\).
02

Determine the magnetic flux for the smaller wire loop

Now, it is needed to compute the magnetic flux through the smaller loop. The magnetic field from the solenoid is constant all over the loop. The area of the loop is \(A = \pi *(0.05)^2 m^2\) because the loop has a diameter of \(10 cm = 0.1 m\). The magnetic flux \(Φ = B * A * cos θ\), where \(θ\) is the angle between the field line and the area vector. As the magnetic field is perpendicular to the plane of the loop, \( θ = 0\) and \( cos θ = 1\). So \(Φ = B * A\).
03

Compute the loop current for the smaller loop

Then, using Faraday's law of electromagnetic induction, which states that the electromotive force \(ε\) around a closed path is equal to the negative rate of change of the magnetic flux enclosed by the path, we compute the electromotive force acting on the loop. In the formula, \(ε = -dΦ/dt\), \(ε\) is the induced emf, and \(Φ\) is the magnetic flux. Because the solenoid's current increase is constant, the change in flux \(dΦ/dt = B * dA/dt + A * dB/dt\), simplifies to \(dΦ/dt = A * dB/dt = A * μ0 * n * dI/dt\) because \(B = μ_0 * n * dI/dt\). The magnitude of \(ε = B * A\). To get the current in the loop, use Ohm's law \(I = ε/R\), where \(R\) is the resistance of the loop.
04

Repeat for the Larger loop Outside the Solenoid

Now, for the second loop, the procedure is almost similar. However, it is important to note that the entire magnetic field of the solenoid goes through the inner section of the loop. Outside the solenoid, the magnetic field is approximately zero. Thus the effective area for this loop is the area of the solenoid 'face', which is \(π * (0.075)^2 m^2\), since the diameter of the solenoid is \(15 cm = 0.15 m\). Perform the same calculations as in steps 2 and 3 using this effective area for this loop.

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