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Chapter 28: Problem 31

A long, straight wire lying along the \(x\) -axis carries a current, \(i\), flowing in the positive \(x\) -direction. A second long, straight wire lies along the \(y\) -axis and has a current \(i\) in the positive \(y\) -direction. What is the magnitude and the direction of the magnetic field at point \(z=b\) on the \(z\) -axis?

Short Answer

Expert verified
Question: Calculate the magnitude and direction of the net magnetic field at a point on the z-axis due to two long, straight wires carrying current along the x-axis and y-axis. Answer: The net magnetic field magnitude at point z = b on the z-axis is \(\sqrt{2} \frac{\mu_0 i}{4\pi b}\), and its direction is at an angle of -45° from the positive x-axis in the xy-plane.

Step by step solution

01

Identify the formula to use

In this case, we use the Biot-Savart Law to find the magnetic field at a specific point due to a long, straight wire. The formula is given by (for a segment of length dl on a wire carrying current i): $$dB = \frac{\mu_0 i}{4\pi}\frac{dl\times \hat r}{r^2}$$ where - dB is the magnetic field at a particular point due to the given segment, - \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \,Tm/A\)), - i is the current in the wire, - dl is the infinitesimal segment of the wire, - \(\hat r\) is the unit vector pointing from the segment to the point where the magnetic field is being calculated, and - r is the distance between the segment and the point.
02

Calculate the magnetic field due to the wire on the x-axis

The magnetic field at a point on the z-axis (z = b) due to a small segment, dx, of the wire on the x-axis can be found using the Biot-Savart law. Integrate the magnetic field over the entire x-axis to find the total magnetic field at the point: $$B_x = \int dB_x = \int_{-\infty}^{\infty} \frac{\mu_0 i}{4\pi}\frac{dx\times \hat r_x}{r_x^2}$$ Since the wire is along the x-axis, the magnetic field's magnitude only has a y-component (and no x- or z-components), and the formula becomes: $$B_x = \int_{-\infty}^{\infty} \frac{\mu_0 i}{4\pi}\frac{b}{(x^2+b^2)^{3/2}} dx$$ Solve the integral to find \(B_x\): $$B_x = \frac{\mu_0 i}{4\pi b}$$
03

Calculate the magnetic field due to the wire on the y-axis

Similarly, find the magnetic field at the point on the z-axis (z = b) due to a small segment, dy, of the wire on the y-axis using the Biot-Savart law. Integrate the magnetic field over the entire y-axis to find the total magnetic field at the point: $$B_y = \int dB_y = \int_{-\infty}^{\infty} \frac{\mu_0 i}{4\pi}\frac{dy\times \hat r_y}{r_y^2}$$ The magnetic field's magnitude only has an x-component in this case (and no y- or z-components), and the formula becomes: $$B_y = \int_{-\infty}^{\infty} \frac{\mu_0 i}{4\pi}\frac{b}{(y^2+b^2)^{3/2}} dy$$ Solve the integral to find: $$B_y = \frac{\mu_0 i}{4\pi b}$$
04

Determine the resultant magnetic field

Now find the net magnetic field at the point on the z-axis. The magnetic field due to the wire on the x-axis is in the negative y-direction, while the magnetic field due to the wire on the y-axis is in the positive x-direction. Therefore, the net magnetic field is given by: $$B = \sqrt{B_x^2 + B_y^2}$$ Substitute the values for \(B_x\) and \(B_y\) and simplify: $$B = \sqrt{\left(\frac{\mu_0 i}{4\pi b}\right)^2 + \left(\frac{\mu_0 i}{4\pi b}\right)^2} \implies B = \sqrt{2} \frac{\mu_0 i}{4\pi b}$$
05

Determine the direction of the magnetic field

To find the direction of the net magnetic field, we can use the tangent of the angle formed by the magnetic field vector in the xy-plane. Since \(B_x\) (negative y-direction) and \(B_y\) (positive x-direction) are equal in magnitude, the angle θ is given by: $$tan(\theta) = \frac{B_y}{B_x} = -1$$ Solving for θ, we get: $$\theta = -45^\circ$$ So the net magnetic field at point z = b on the z-axis has a magnitude of \(\sqrt{2} \frac{\mu_0 i}{4\pi b}\) and is directed at an angle of -45° from the positive x-axis in the xy-plane.

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Most popular questions from this chapter

A toy airplane of mass \(0.175 \mathrm{~kg}\), with a charge of \(36 \mathrm{mC}\), is flying at a speed of \(2.8 \mathrm{~m} / \mathrm{s}\) at a height of \(17.2 \mathrm{~cm}\) above and parallel to a wire, which is carrying a 25 - A current; the airplane experiences some acceleration. Determine this acceleration.

Consider an electron to be a uniformly dense sphere of charge, with a total charge of \(-e=-1.60 \cdot 10^{-19} \mathrm{C}\) spinning at an angular frequency, \(\omega\). a) Write an expression for its classical angular momentum of rotation, \(L\) b) Write an expression for its magnetic dipole moment, \(\mu\). c) Find the ratio, \(\gamma_{e}=\mu / L\), known as the gyromagnetic ratio.

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28.39 The figure shows a cross section across the diameter of a long, solid, cylindrical conductor. The radius of the cylinder is \(R=10.0 \mathrm{~cm} .\) A current of \(1.35 \mathrm{~A}\) is uniformly distributed through the conductor and is flowing out of the page. Calculate the direction and the magnitude of the magnetic field at positions \(r_{\mathrm{a}}=0.0 \mathrm{~cm}\) \(r_{\mathrm{b}}=4.00 \mathrm{~cm}, r_{\mathrm{c}}=10.00 \mathrm{~cm}\) and \(r_{d}=16.0 \mathrm{~cm} .\)
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