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Chapter 7: Problem 15

A long straight wire carrying a current \(I\) lies in the same plane as a rectangular loop with sides of length \(a\) and \(b\). The sides \(a\) are parallel to the wite and distances \(d\) and \(d+b\) from it. What is the total magnetic flux linking the loop?

Short Answer

Expert verified
Answer: The total magnetic flux through the rectangular loop is given by: \(\Phi_{total} = \frac{\mu_0 Ia}{2\pi} \left(\frac{1}{d} +\frac{1}{d+b}\right)\) where \(\mu_0\) is the permeability of free space, I is the current in the wire, a is the side length of the loop parallel to the wire, d is the distance between the wire and the loop, and b is the side length of the loop perpendicular to the wire.

Step by step solution

01

Determine the magnetic field at a distance r from the wire

Let's find the expression for the magnetic field at a distance r from the long straight wire carrying a current I using the Biot-Savart Law. The magnetic field (B) is given by: \(B = \frac{\mu_0 I}{2\pi r}\) Where \(\mu_0\) is the permeability of free space, and r is the distance from the wire.
02

Find the magnetic field along the sides of the loop

Now we need to find the magnetic fields along the two sides of length a parallel to the wire. The distances to these sides are d and d+b. So, the magnetic fields are: \(B_1 = \frac{\mu_0 I}{2\pi d}\) and \(B_2 = \frac{\mu_0 I}{2\pi (d+b)}\)
03

Calculate the flux through each side of the loop

To find the total magnetic flux, we need to calculate the flux through each side of the loop. The flux through a side with a magnetic field B and side length a is given by the product of B, the side length, and the angle between B and the side. Since the magnetic field is perpendicular to the sides of length a, the angle between B and the side is 0 degrees, so the cos(angle) is 1. The flux through each side is: \(\Phi_1 = B_1 \cdot a \cdot \cos(0) = \frac{\mu_0 I}{2\pi d} \cdot a\) \(\Phi_2 = B_2 \cdot a \cdot \cos(0) = \frac{\mu_0 I}{2\pi (d+b)} \cdot a\)
04

Calculate the total magnetic flux through the loop

To find the total magnetic flux through the loop, we need to sum the fluxes through each side: \(\Phi_{total} = \Phi_1 + \Phi_2\) Substitute the expressions for \(\Phi_1\) and \(\Phi_2\) from Step 3: \(\Phi_{total} = \frac{\mu_0 I}{2\pi d} \cdot a +\frac{\mu_0 I}{2\pi (d+b)} \cdot a\) Factor out the common terms: \(\Phi_{total} = \frac{\mu_0 Ia}{2\pi} \left(\frac{1}{d} +\frac{1}{d+b}\right)\) This is the total magnetic flux linking the rectangular loop.

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Most popular questions from this chapter

Photoclectrons are liberated from a plate by ultraviolet radiation, their initial velocity being negligible. A magnetic field \(\mathbf{B}\) is maintained parallel to the plate and an E-field perpendicular to it. The E-field is produced by a second plate parallel to the first, a distance \(d\) from it and at a positive potential \(V\) with respect to it. Show that the value of \(d\) for which the current just fails to pass between the plates is \(\left(2 m, V / e B^{2}\right)^{1 / 2}\), where e and \(m_{e}\) are the charge and mass of the clectron.

Modern magnetometers for measuring B-fields are described in Comment C12.2. In the older vibration magnetometer, a small permanent dipole was allowed to perform small angular oscillations in a horizontal plane first in a standard field \(\mathbf{B}_{0}\) and then in the unknown \(\mathrm{B}\). If the periods of oscillation were, respectively, \(T_{0}\) and \(T\), show that \(B=B_{0} T_{6}^{2} T^{2}\),

Take a small rectangular current loop of sides \(\mathrm{d} x\), dy situated at the origin in the xy plane and carrying a current \(I\) round its four sides. Find the B-field at a. field point P a great distance from the loop along the \(x\) axis. Show that the field has the form \(\mu_{0} m /\left(4 \pi x^{3}\right)\), where \(m\) is its dipole moment. (Note that all four sides produce a field.)

A long thin metal strip of width \(2 L\) carries a current along its length. The current is distributed uniformly across the width with a surface density \(J\), (Sec. 1.7). Find the B-ficld (a) at a point in the plane of the strip at a distance \(R(>L)\) from its centre and (b) at a point a perpendicular distance \(R\) from its centre. Check that (a) yields \((7.19)\) as \(L / R \rightarrow 0\). What does (b) become if the sheet is of infinite width?

A wire is bent into a circle of radius \(a\) and a total charge \(Q\) is distributed uniformly around it. Find the B-field at the centre if the wire rotates with an angular velocity \(\omega\) about the axis of the circle.
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