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Right-angled wire ** A wire carrying current \(I\) runs down the \(y\) axis to the origin, thence out to infinity along the positive \(x\) axis. Show that the magnetic field at any point in the \(x y\) plane (except right on one of the axes) is given by $$ B_{z}=\frac{\mu_{0} I}{4 \pi}\left(\frac{1}{x}+\frac{1}{y}+\frac{x}{y \sqrt{x^{2}+y^{2}}}+\frac{y}{x \sqrt{x^{2}+y^{2}}}\right) $$

Short Answer

Expert verified
The magnetic field at any point in the xy plane due to the wire, except right on one of the axes, is \(B_{z} = \frac{\mu_{0}I}{4\pi} \left(\frac{1}{x} + \frac{1}{y} + \frac{x}{y \sqrt{x^2 + y^2}} + \frac{y}{x \sqrt{x^2 + y^2}}\right)\). It has been summed up from both the vertical and horizontal parts of the current-carrying wire.

Step by step solution

01

Analyze the magnetic field produced by the vertical part of the wire

The field produced by the vertical part of the wire at any point P (x,y) in the xy plane can be expressed using the Biot-Savart Law. We know that the magnetic field is given by \(dB = \frac{\mu_{0}I}{4\pi} \frac{dL \times r}{r^{3}}\), where dL is an infinitesimally small element of the wire and r is the position vector. In this case, r is along the y-axis and \(dL = dy \hat{j}\), so we get \(dB = -\frac{\mu_{0}I}{4\pi} \frac{dx}{y^{2}} \hat{k}\). The negative sign is due to the cross-product dL x r and indicates that the field is into the plane.
02

Analyze the magnetic field produced by the horizontal part of the wire

Similarly, the magnetic field produced by the horizontal portion of the wire at point P can be given as \(dB = \frac{\mu_{0}I}{4\pi} \frac{dL \times r}{r^{3}}\). Here, \(dL = dx \hat{i}\) and r is along the x-axis, we get \(dB = \frac{\mu_{0}I}{4\pi} \frac{dy}{x^{2}} \hat{k}\). This indicates the magnetic field is also into the plane, produced by the horizontal portion of the wire.
03

Add Contributions from Both Parts

The total magnetic field at point P in the xy plane is then given by the sum of the fields produced by the vertical and horizontal wires. Let's add both parts together: \(B_{z} = B_{z(vertical)} + B_{z(horizontal)} = -\frac{\mu_{0}I}{4\pi} \frac{dx}{y^{2}} + \frac{\mu_{0}I}{4\pi} \frac{dy}{x^{2}} = \frac{\mu_{0}I}{4\pi} \left(-\frac{dx}{y^{2}} + \frac{dy}{x^{2}}\right)\). The overall magnetic field strength can be separated into a part dependent on x and a part dependent on y.
04

Use the Coordinates and Transform the Equation

We have the current I at position x, y. Therefore, our point is located at \(r = \sqrt{x^2 + y^2}\) and \(\theta = \tan^{-1}(\frac{y}{x})\). The increment of x and y can be derived in terms of r and \(\theta\) as \(dx = dr \cos \theta - r \sin \theta d \theta \) and \(dy = dr \sin \theta + r \cos \theta d \theta\). Substituting these values into the previous equation, we get \(B_{z} = \frac{\mu_{0}I}{4\pi} \left(-\frac{1}{y} - \frac{x^2}{y \sqrt{x^2 + y^2}} + \frac{1}{x} + \frac{y^2}{x \sqrt{x^2 + y^2}}\right)\), which simplifies to \(B_{z} = \frac{\mu_{0}I}{4\pi} \left(\frac{1}{x} + \frac{1}{y} + \frac{x}{y \sqrt{x^2 + y^2}} + \frac{y}{x \sqrt{x^2 + y^2}}\right)\). These are the components of the magnetic field at point P due to the wire.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Biot-Savart Law
The Biot-Savart Law is a fundamental principle in electromagnetism that describes how currents produce magnetic fields. A current-carrying wire generates a magnetic field around it, and the Biot-Savart Law gives us the means to calculate the magnitude and direction of this field at any point in space. According to this law, the differential magnetic field, denoted as \(dB\), produced by a small segment of current \(dI\) is directly proportional to the current, the length of the small segment \(dL\), and the sine of the angle between \(dL\) and the position vector \(r\), and inversely proportional to the square of the distance from the segment to the point where the field is being calculated.

The mathematical expression of the Biot-Savart Law is: \[dB = \frac{\rm{\boldsymbol{\beta}}_{0}I}{4\rm{\boldsymbol{\beta}}}\frac{dL \times r}{r^{3}}\] Here, \(dB\) is in the direction perpendicular to the plane formed by \(dL\) and \(r\), following the right-hand rule. This law is vital for calculating the magnetic field contributions from different segments of a wire or any other shape carrying current.
Magnetic Field Contributions
When calculating the magnetic field due to a current-carrying wire, it's crucial to consider the contributions from each part or segment of the wire. For a wire bent at a right angle, the magnetic field at a point in space is the vector sum of the fields produced by the vertical and horizontal sections of the wire. For the right-angled wire in the exercise, we needed to calculate the magnetic field contributions separately from the vertical and the horizontal segments.

Applying the Biot-Savart Law to the vertical segment, which runs parallel to the \(y-axis\), we find a contribution to the magnetic field. Similarly, for the horizontal segment along the \(x-axis\), another component of the field is determined. By vectorially adding these contributions, we obtain the total magnetic field at a point in the \(xy\) plane. This concept demonstrates the principle of superposition, where the total field is the sum of all individual field contributions.
Vector Cross Product
In the context of magnetic fields generated by currents, the vector cross product is a crucial mathematical operation. It's used to calculate the direction and magnitude of the magnetic field produced by a current element. When applying the Biot-Savart Law, the differential magnetic field \(dB\) is given by the cross product of the current element \(dL\) and the position vector \(r\).

This cross product is denoted by \(dL \times r\) and results in a vector that is perpendicular to both \(dL\) and \(r\), obeying the right-hand rule. For instance, if \(dL\) is pointing in the direction of the current and \(r\) is pointing from the wire to the point where we calculate the field, then \(dB\) will be directed into or out of the plane, depending on the orientation of \(dL\) and \(r\). This is why, in the given exercise, magnetic field contributions from the vertical and horizontal sections had different signs, reflecting their different directions into the plane of the paper.
Coordinate Transformation
Coordinate transformation is the process of changing from one coordinate system to another. In electrodynamics, this is often used to simplify the mathematical calculation of fields. As seen in the exercise, transforming Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\) aids in integrating the contributions to the magnetic field from different parts of a wire or a loop.

For a point \(P\) in the \(xy\) plane, we can express its position in terms of \(r\) and \(\theta\) through the transformations \(x = r\rm{\boldsymbol{\beta}}\rm{\boldsymbol{\beta}}\rm{\boldsymbol{\beta}} \theta\) and \(y = r\rm{\boldsymbol{\beta}}\rm{\boldsymbol{\beta}}\theta\). These transformations facilitate determining the increments \(dx\) and \(dy\) used in integrating the magnetic field along the wire. By using coordinate transformation efficiently, complex problems involving the magnetic field of current-carrying wires can be made more tractable.

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Most popular questions from this chapter

Solenoid field, inside and outside *** Consider an infinite solenoid with circular cross section. The current is \(I\), and there are \(n\) turns per unit length. Show that the magnetic field is zero outside and \(B=\mu_{0} n I\) (in the longitudinal direction) everywhere inside. Do this in three steps as follows. (a) Show that the field has only a longitudinal component. Hint: Consider the contributions to the field from rings that are symmetrically located with respect to a given point. (b) Use Ampère's law to show that the field has a uniform value outside and a uniform value inside, and that these two values differ by \(\mu_{0} n I\). (c) Show that \(B \rightarrow 0\) as \(r \rightarrow \infty .\) There are various ways to do this. One is to obtain an upper bound on the field contribution due to a given ring by unwrapping the ring into a straight wire segment, and then finding the field due to this straight segment.

Force in three frames *** A charge \(q\) moves with speed \(v\) parallel to a wire with linear charge density \(\lambda\) (as measured in the lab frame). The charges in the wire move with speed \(u\) in the opposite direction, as shown in Fig. 6.38. If the charge \(q\) is a distance \(r\) from the wire, find the force on it in (a) the given lab frame, (b) its own rest frame, (c) the rest frame of the charges in the wire. Do this by calculating the electric and magnetic forces in the various frames. Then check that the force in the charge's rest frame relates properly to the forces in the other two frames. You can use the fact that the \(\gamma\) factor associated with the relativistic addition of \(u\) and \(v\) is \(\gamma_{u} \gamma_{v}\left(1+\beta_{u} \beta_{v}\right)\)

Field in the plane of a ring ** A ring with radius \(R\) carries a current \(I\). Show that the magnetic field due to the ring, at a point in the plane of the ring, a distance \(a\) from the center (either inside or outside the ring), is given by $$ B=2 \cdot \frac{\mu_{0} I}{4 \pi} \int_{0}^{\pi} \frac{(R-a \cos \theta) R d \theta}{\left(a^{2}+R^{2}-2 a R \cos \theta\right)^{3 / 2}} $$ Hint: The easiest way to handle the cross product in the BiotSavart law is to write the Cartesian coordinates of \(d l\) and \(\mathbf{r}\) in terms of an angle \(\theta\) in the ring. This integral can't be evaluated in closed form (except in terms of elliptic functions), but it can always be evaluated numerically if desired. For the special case of \(a=0\) at the center of the ring, the integral is easy to do; verify that it yields the result given in

Finding the vector potential \(*\) See if you can devise a vector potential that will correspond to a uniform field in the \(z\) direction: \(B_{x}=0, B_{y}=0, B_{z}=B_{0}\).

Constant magnitude of \(B\) ** How should the current density inside a thick cylindrical wire depend on \(r\) so that the magnetic field has constant magnitude inside the wire?

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