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Chapter 27: Problem 45

Figure \(\mathbf{E} 27 . \mathbf{4 5}\) shows a portion of a silver ribbon with \(z_{1}=11.8 \mathrm{~mm}\) and \(y_{1}=0.23 \mathrm{~mm},\) carrying a current of \(120 \mathrm{~A}\) in the \(+x\) -direction. The ribbon lies in a uniform magnetic field, in the \(y-\) direction, with magnitude 0.95 T. Apply the simplified model of the Hall effect presented in Section \(27.9 .\) If there are \(5.85 \times 10^{28}\) free electrons per cubic meter, find (a) the magnitude of the drift velocity of the electrons in the \(x\) -direction; (b) the magnitude and direction of the electric field in the \(z\) -direction due to the Hall effect; (c) the Hall emf.

Short Answer

Expert verified
The magnitude of the drift velocity of the electrons in the x direction is calculated in step 1, the magnitude and direction of the electric field in the z direction due to the Hall effect is calculated in step 2, and the Hall emf is determined in step 3 using the equations outlined in the detailed solution steps.

Step by step solution

01

Find the drift velocity

First, use the given current \(I\), the given density of free electrons \(n\), cross-sectional area of the ribbon \(A\) and the charge of an electron to find the drift velocity \(v\). The cross-sectional area of the ribbon can be determined by multiplying the thickness of the ribbon, \(z_{1}\), by the width of the ribbon, \(y_{1}\). Afterwards use the formula: \(I = nAvq\), to solve for \(v\).
02

Calculate the Hall Electric Field

Use the simplified model of the Hall effect, the formula for the electric field due to the Hall effect, \(E = Bvd\), where \(B\) is the magnetic field and \(v_d\) is the drift velocity. Use the found drift velocity and the given magnetic field to calculate the electric field.
03

Calculate the Hall emf

Finally, to find the Hall emf \(V_H\), use the formula \(V_H = Ed\), where \(E\) is the Hall electric field and \(d\) is the distance over which this field applies. In this case, \(d\) is the thickness of the silver ribbon. Use the calculated electric field and the given thickness of the ribbon to find the Hall emf. This is the final step and will yield the solution to part (c) of the question.

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Key Concepts

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

Drift Velocity
The drift velocity is a concept closely related to electric current. It refers to the average velocity of free charges, such as electrons, as they move through a conductor due to an applied electric field. In general, electrons move randomly at high speeds, but when a current is applied, they gain a small net speed in the direction of the current 鈥� this is their drift velocity.

To calculate drift velocity (\(v_d\)), you need information about the current (\(I\)), the number of free electrons per unit volume (\(n\)), and the cross-sectional area (\(A\)) of the conductor. The formula used is:
  • \[ I = nAv_eq \]
  • \(v_d = \frac{I}{nAq} \)
where \(q\) is the charge of an electron (~\(1.6 \times 10^{-19} C\)). For the given exercise, by using the ribbon's thickness and width, you can find its cross-sectional area and then solve for drift velocity.
Magnetic Field
A magnetic field is a vector field that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials. It is typically represented by the symbol \(B\) and measured in teslas (T).

In the Hall effect, a magnetic field imposes a force on the moving charges within the conductor, pushing them perpendicular to both the magnetic field and the direction of current. This force is what causes the Hall voltage to develop across the material. The magnitude of this force can be calculated using the formula:
  • \[ F = q(v_d \times B) \]
where \(v_d\) is the drift velocity and \(B\) is the magnetic field. The vector nature of the magnetic field means that it鈥檚 important to consider the direction when calculating effects like the Hall voltage. In this exercise, the magnetic field is given as pointing in the \(y\)-direction, affecting charges moving in the \(x\)-direction.
Electric Field
An electric field represents the force per unit charge exerted on a charged particle. It鈥檚 a crucial concept in understanding the Hall effect, which involves the creation of an electric field perpendicular to both the current and the magnetic field. The Hall effect demonstrates how an imbalance of charge can lead to a transverse electric field. This field causes the Hall voltage across the material.

For the Hall effect, use the formula:
  • \[ E = Bv_d \]
where \(E\) is the electric field, \(B\) is the magnitude of the magnetic field, and \(v_d\) is the drift velocity. The direction of this electric field is determined by the directions of both the current and the magnetic field in the conductor. Given that the magnetic field and drift velocity are perpendicular, the resulting electric field is also orthogonal to both. In this scenario, this field lies in the \(z\)-direction.

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Most popular questions from this chapter

A dc motor with its rotor and field coils connected in series has an internal resistance of \(3.2 \Omega .\) When the motor is running at full load on a \(120 \mathrm{~V}\) line, the emf in the rotor is \(105 \mathrm{~V}\). (a) What is the current drawn by the motor from the line? (b) What is the power delivered to the motor? (c) What is the mechanical power developed by the motor?

A horizontal rectangular surface has dimensions \(2.80 \mathrm{~cm}\) by \(3.20 \mathrm{~cm}\) and is in a uniform magnetic field that is directed at an angle of \(30.0^{\circ}\) above the horizontal. What must the magnitude of the magnetic field be to produce a flux of \(3.10 \times 10^{-4} \mathrm{~Wb}\) through the surface?

The large magnetic fields used in MRI can produce forces on electric currents within the human body. This effect has been proposed as a possible method for imaging "biocurrents" flowing in the body, such as the current that flows in individual nerves. For a magnetic field strength of \(2 \mathrm{~T}\), estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical nerve that has a diameter of \(1.5 \mathrm{~mm} .\) Assume that the entire nerve carries a current due to an applied voltage of \(100 \mathrm{mV}\) (that of a typical action potential). The resistivity of the nerve is \(0.6 \Omega \cdot \mathrm{m}\). (a) \(6 \times 10^{-7} \mathrm{~N} ;\) (b) \(1 \times 10^{-6} \mathrm{~N} ;\) (c) \(3 \times 10^{-4} \mathrm{~N}\) (d) \(0.3 \mathrm{~N}\).

Magnetic Moment of the Hydrogen Atom. In the Bohr model of the hydrogen atom (see Section 39.3 ), in the lowest energy state the electron orbits the proton at a speed of \(2.2 \times 10^{6} \mathrm{~m} / \mathrm{s}\) in a circular orbit of radius \(5.3 \times 10^{-11} \mathrm{~m}\). (a) What is the orbital period of the electron? (b) If the orbiting electron is considered to be a current loop, what is the current \(I ?\) (c) What is the magnetic moment of the atom due to the motion of the electron?

A flat, square surface with side length \(3.40 \mathrm{~cm}\) is in the \(x y\) -plane at \(z=0 .\) Calculate the magnitude of the flux through this surface produced by a magnetic field \(\overrightarrow{\boldsymbol{B}}=(0.200 \mathrm{~T}) \hat{\imath}+(0.300 \mathrm{~T}) \hat{\jmath}-(0.500 \mathrm{~T}) \hat{\boldsymbol{k}}\).
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