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Force on a Current Loop in a Nonuniform Magnetic Field. It was shown in Section 27.7 that the net force on a current loop in a uniform magnetic field is zero. But what if \(\vec { B }\) is not uniform? Figure P27.85 shows a square loop of wire that lies in the \(x y\) -plane. The loop has corners at \(( 0,0 ) , ( 0 , L ) , ( L , 0 ) ,\) and \(( L , L )\) and carries a constant current \(I\) in the clockwise direction. The magnetic field has no \(x\) -component but has both \(y\) - and z-components: \(\vec { B } = \left( B _ { 0 } z / L \right) \hat { J } + \left( B _ { 0 } y / L \right) \hat { k } ,\) where \(B _ { 0 }\) is a positive constant. (a) Sketch the magnetic field lines in the \(y z\) -plane. (b) Find the magnitude and direction of the magnetic force exerted on each of the sides of the loop by integrating Eq. (27.20). (c) Find the magnitude and direction of the net magnetic force on the loop.

Short Answer

Expert verified
The net force on the loop is \( IB_0 L \hat{k} \). The direction is in the \( z \)-axis.

Step by step solution

01

Analyzing the Magnetic Field Components

The magnetic field is given by \( \vec{B} = \left( \frac{B_0 z}{L}\right) \hat{j} + \left( \frac{B_0 y}{L} \right) \hat{k} \). This means it has a \( y \)-component increasing linearly with \( z \) and a \( z \)-component increasing linearly with \( y \). There is no \( x \)-component, thus the field lines in the \( yz \)-plane will appear to form straight lines angled depending on \( y \) and \( z \) values.
02

Finding the Force on Side 1 (from (0,0) to (0,L))

The current on this side flows in the \( +y \) direction, thus \( d\vec{l} = \hat{j} dy \). The magnetic field at side 1 has only a \( z \)-component \( \left( B_0 y / L \right) \hat{k} \). Using \[ d\vec{F} = I (d\vec{l} \times \vec{B}) \], the cross product \( \hat{j} \times \left( B_0 y / L \right)\hat{k} = -\left( B_0 y / L \right)\hat{i} \). Thus, \( d\vec{F}_1 = -I \frac{B_0 y}{L} dy \hat{i} \). Integrating over \( y \) from 0 to \( L \), the total force on side 1 is \( \vec{F}_1 = -\frac{IB_0 L}{2} \hat{i} \).
03

Finding the Force on Side 2 (from (0,L) to (L,L))

The current in this side flows in the \( +x \) direction, thus \( d\vec{l} = \hat{i} dx \). The magnetic field here has no \( x \)-component, \( \vec{B} = \left( B_0 \frac{L}{L} \right) \hat{j} = B_0 \hat{j} \). Calculating the cross product, \( \hat{i} \times B_0 \hat{j} = B_0 \hat{k} \). \( d\vec{F}_2 = IB_0 dx \hat{k} \). Integrating over \( x \) from 0 to \( L \), \( \vec{F}_2 = IB_0 L \hat{k} \).
04

Finding the Force on Side 3 (from (L,L) to (L,0))

The current here flows in the \( -y \) direction, so \( d\vec{l} = -\hat{j} dy \). The magnetic field at side 3 is \( \vec{B} = \left( B_0 \frac{L}{L} \right) \hat{j} + \left( B_0 y / L \right) \hat{k} = B_0 \hat{j} + \left( B_0 y / L \right) \hat{k} \). The cross product \( -\hat{j} \times ( B_0 \hat{j} + \frac{B_0 y}{L} \hat{k}) = -\frac{B_0^2 y}{L} \hat{i} \). Integrating over \( y \), \( \vec{F}_3 = \int_{L}^0 -\frac{IB_0 y}{L} dy \hat{i} = \frac{IB_0 L}{2} \hat{i} \).
05

Finding the Force on Side 4 (from (L,0) to (0,0))

The current flows in the \( -x \) direction, meaning \( d\vec{l} = -\hat{i} dx \). The magnetic field \( \vec{B} = 0 \) (since \( z = 0\), reducing \( \( B_0 z / L\) \hat{j} = 0 \). So the force \( d\vec{F}_4 = 0 \). Thus, \( \vec{F}_4 = 0 \).
06

Calculating the Net Magnetic Force

Summing the forces from all sides: \( \vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \vec{F}_4 = -\frac{IB_0 L}{2} \hat{i} + IB_0 L \hat{k} + \frac{IB_0 L}{2} \hat{i} = IB_0 L \hat{k} \). The \( i \)-components cancel, and only the \( k \)-component remains.

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Key Concepts

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

Nonuniform Magnetic Field
When we talk about a magnetic field, we're referring to an invisible field produced by electric charges in motion, and it can vary in strength and direction. A nonuniform magnetic field means that the strength and/or direction of the field changes from one point to another within the space it occupies. In our specific problem, the magnetic field has a dependency on position because its components depend on the coordinates (\(y\) and \(z\)). This positional dependence results in a magnetic field that varies in the \(yz\)-plane, hence it's nonuniform.
These types of magnetic fields require careful mathematical handling to determine forces on objects within them, particularly if the magnetic field components are not the same throughout the region of interest.
  • The \(y\)-component increases linearly with the \(z\) coordinate, described by \(\frac{B_0 z}{L}\), denoting how the field gets stronger with increasing \(z\).
  • The \(z\)-component increases linearly with the \(y\) coordinate, given as \(\frac{B_0 y}{L}\), indicating the field intensifies as \(y\) increases.
Understanding how \(\vec{B}\) changes with position is crucial in calculating the magnetic forces on current-carrying loops or wires that pass through these regions.
Integration in Physics
Integration is a fundamental concept in physics, particularly when dealing with continuous quantities like fields or currents. In our problem, we need to calculate the magnetic force on each section of the loop, which involves a continuous distribution of current. This requires us to sum up tiny contributions of forces along each segment of the wire loop.
Using the formula:\[ d\vec{F} = I \left( d\vec{l} \times \vec{B} \right) \]
We integrate across the path of the current to find the total force on each segment. Here's how integration comes into play:
  • For each segment of the wire, a small length element \(d\vec{l}\) is considered in the direction of the current.
  • By integrating this small force element over the entire length of the wire segment, we obtain the total force exerted by the magnetic field.
This step-wise integration across the loop accounts for changes in the magnetic field along the path, especially important in a nonuniform field, ensuring that the calculation of the force is accurate and representative of the physical situation.
Cross Product in Vectors
The cross product is a vector operation vital in physics, providing information on forces, torque, and more, which are also vector quantities. In our context, the cross product helps us find the direction and magnitude of the magnetic force acting on a current-carrying wire within a magnetic field.
The formula for the magnetic force involves a cross product:\[ d\vec{F} = I \left( d\vec{l} \times \vec{B} \right) \]
Here,
  • \(d\vec{l}\) denotes the differential element of the wire in the direction of current flow.
  • \(\vec{B}\) represents the magnetic field vector at the point where the wire segment is located.
The cross product calculates a new vector that is perpendicular to the plane formed by \(d\vec{l}\) and \(\vec{B}\). The magnitude of this vector (and thus the force) is given by:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(\theta) \]
Where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\). The direction of the resultant vector follows the right-hand rule, a handy tool to find out this orientation: point your fingers in the direction of \(\vec{a}\) and curl them towards \(\vec{b}\), your thumb will point in the direction of the resulting vector.
Understanding the cross product ensures proper calculation of the magnetic force's direction and magnitude on segments within a nonuniform magnetic field.

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Most popular questions from this chapter

Paleoclimate. Climatologists can determine the past temperature of the earth by comparing the ratio of the isotope oxygen-18 to the isotope oxygen-16 in air trapped in ancient ice sheets, such as those in Greenland. In one method for separating these isotopes, a sample containing both of them is first singly ionized (one electron is removed) and then accelerated from rest through a potential difference \(V\) . This beam then enters a magnetic field \(B\) at right angles to the field and is bent into a quarter-circle. A particle detector at the end of the path measures the amount of each isotope. (a) Show that the separation \(\Delta r\) of the two isotopes at the detector is given by $$\Delta r = \frac { \sqrt { 2 e V } } { e B } \left( \sqrt { m _ { 18 } } - \sqrt { m _ { 16 } } \right)$$ where \(m _ { 16 }\) and \(m _ { 18 }\) are the masses of the two oxygen isotopes, (b) The measured masses of the two isotopes are \(2.66 \times\) \(10 ^ { - 26 } \mathrm { kg } \left( ^ { 16 } \mathrm { O } \right)\) and \(2.99 \times 10 ^ { - 26 } \mathrm { kg } ( 8 \mathrm { O } ) .\) If the magnetic field is \(0.050 \mathrm { T } ,\) what must be the accelerating potential \(V\) so that these two isotopes will be separated by 4.00\(\mathrm { cm }\) at the detector?

A particle with charge 7.80\(\mu \mathrm{C}\) is moving with velocity \(\vec{\boldsymbol{v}}=-\left(3.80 \times 10^{3} \mathrm{m} / \mathrm{s}\right) \hat{\boldsymbol{J}}\) . The magnetic force on the particle is measured to be \(\vec{\boldsymbol{F}}=+\left(7.60 \times 10^{-3} \mathrm{N}\right) \hat{\boldsymbol{\imath}}-\left(5.20 \times 10^{-3} \mathrm{N}\right) \hat{\boldsymbol{k}}\) (a) Calculate all the components of the magnetic field you can from this information. (b) Are there components of the magnetic field that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\vec{B} \cdot \vec{F} .\) What is the angle between \(\vec{B}\) and \(\vec{\boldsymbol{F}} ?\)

A singly ionized (one electron removed) \(^{40} \mathrm{K}\) atom passes through a velocity selector consisting of uniform perpendicular electric and magnetic fields. The selector is adjusted to allow ions having a speed of 4.50 \(\mathrm{km} / \mathrm{s}\) to pass through undeflected when the magnetic field is 0.0250 T. The ions next enter a second uniform magnetic field \(\left(B^{\prime}\right)\) oriented at right angles to their velocity. 40 contains 19 protons and 21 neutrons and has a mass of \(6.64 \times 10^{-26} \mathrm{kg} .\) (a) What is the magnitude of the electric field in the velocity selector? (b) What must be the magnitude of \(B^{\prime}\) so that the ions will be bent into a semicircle of radius 12.5 \(\mathrm{cm} ?\)

The Electromagnetic Pump. Magnetic forces acting on conducting fluids provide a convenient means of pumping these fluids. For example, this method can be used to pump blood without the damage to the cells that can be caused by a mechanical pump. A horizontal tube with rectangular cross section (height \(h ,\) width \(w )\) is placed at right angles to a uniform magnetic field with magnitude \(B\) so that a length \(l\) is in the field (Fig. P27.90). The tube is filled with a conducting liquid, and an electric current of density \(J\) is maintained in the third mutually perpendicular direction. (a) Show that the difference of pressure between a point in the liquid on a vertical plane through \(a b\) and a point in the liquid on another vertical plane through \(c d ,\) under conditions in which the liquid is prevented from flowing, is \(\Delta p = J / B\) . (b) What current density is needed to provide a pressure difference of 1.00 atm between these two points if \(B = 2.20 \mathrm { T }\) and \(l = 35.0 \mathrm { mm } ?\)

A magnetic field exerts a torque \(\tau\) on a round current carrying loop of wire. What will be the torque on this loop (in terms of \(\tau\) if its diameter is tripled?

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