Chapter 19, Problem 33E

(a)

To determine

The magnitude and direction of the magnetic force on the electrons in the conductor shown in Figure 19.37.

Answer to Problem 33E

The magnitude of the force is $\underset{\_}{e{v}_{d}B}$, acting upward.

Explanation of Solution

The magnitude of magnetic force experienced on a moving charge can be written as,

    $F=\left|q\right|vB\sin \theta$        (I)

Here, $F$ is the magnitude of the magnetic force, $q$ is the charge of the particle, $v$ is the velocity of the charge, $B$ is the magnetic field, and $\theta$ is the angle between magnetic field and velocity of the charge. Consider a straight segment of wire which is part of a closed circuit carrying current.

Let $v_d$ represent the drift velocity of the electrons in the conductor.

Conclusion:

Substitute $-e$ for $q$, $v_d$ for $v$, and $90°$ for $\theta$ in equation (I) to find the $F$.

    $\begin{array}{c}F=\left|-e\right|v_{d}B\sin 90°\\ =e{v}_{d}B\end{array}$        (II)

Direction of force will be in the direction of the cross product $\left(v\times B\right)$. Since the electron is negatively charged the direction will be reversed or it will in the direction of the cross product $\left(B\times v\right)$. That is upward.

Therefore, the magnitude of the force is $\underset{\_}{e{v}_{d}B}$, acting upward.

(b)

To determine

The magnitude and direction of the electric field that would exert an equal but opposite force on the electrons.

Answer to Problem 33E

The magnitude of electric field that would exert an equal but opposite force on the electrons is $v_{d}B$, and it is directed upward.

Explanation of Solution

Write the expression to find the force on a charge in an electric field.

    $F=qE$        (III)

Here, $F$ is the force, $q$ is the charge of the particle, and $E$ is the electric field.

Rewrite equation (III) to find $E$.

    $E=\frac{F}{\left|q\right|}$        (VI)

For an electron, $q$ is $-e$. So equation (III) becomes,

    $E=\frac{F}{e}$        (V)

Substitute equation (II) in part (a) in (V) to find $E$.

    $\begin{array}{c}E=\frac{e{v}_{d}B}{e}\\ =v_{d}B\end{array}$

The direction of electric force will be same as that of the force. That is upward.

Conclusion:

Therefore, the magnitude of electric field that would exert an equal but opposite force on the electrons is $v_{d}B$, and it is directed upward.

(c)

To determine

The potential difference across the conductor to produce the electric field.

Answer to Problem 33E

Potential difference across the conductor to produce the electric field is $\underset{\_}{a{v}_{d}B}$.

Explanation of Solution

Write the expression to find the work done on a charge in a potential difference.

    $W=q\Delta V$        (VI)

Here, $W$ is the work done, and $\Delta V$ is the change in potential difference.

Write the expression to find the work done.

    $W=Fd$        (VII)

Here, $d$ is the distance.

Substitute equations (III) and (VI) in (VII) to and solve for $\Delta V$.

    $\begin{array}{l}q\Delta V=qEd\\ \Delta V=Ed\end{array}$        (VIII)

Conclusion:

From part (a) it is found that the magnitude of electric field that would exert an equal but opposite force on the electrons is $v_{d}B$.

Substitute $v_{d}B$ for $E$, and $a$ for $d$ in equation (VIII) to find $\Delta V$.

    $\Delta V=a{v}_{d}B$

Therefore, potential difference across the conductor to produce the electric field is $\underset{\_}{a{v}_{d}B}$.

(d)

To determine

The potential difference when no external electric field is applied.

Answer to Problem 33E

The potential difference when no external electric field is applied is $\underset{\_}{a{v}_{d}B}$.

Explanation of Solution

The establishment of a potential difference across a conductor in a magnetic field is called Hall effect.

Write the expression to find the hall voltage.

  $V_H=Bvd$        (IX)

Here, $V_H$ is the Hall voltage.

Conclusion:

Substitute $v_d$ for $v$, and $a$ for $d$ in equation (IX) to find $V_H$.

    $V_H=a{v}_{d}B$        (X)

Therefore, the potential difference when no external electric field is applied  is $\underset{\_}{a{v}_{d}B}$.

(e)

To determine

The Hall potential difference of the circuit.

Answer to Problem 33E

The Hall potential difference is $\underset{\_}{2\times {10}^{-5}\text{ V}}$.

Explanation of Solution

The establishment of a potential difference across a conductor in a magnetic field is called Hall effect. Use equation (X) in part (d) to find the Hall potential difference.

Conclusion:

Substitute $1.0\times {10}^{-3}\text{ m}\cdot {\text{s}}^{-1}$ for $v_d$, and $1.0\times {10}^{-2}\text{ m}$ for $a$, and $2\text{ T}$ for $B$ in equation (X) to find $V_H$.

    $\begin{array}{c}{V}_H=\left(2\text{ T}\right)\left(1.0\times {10}^{-3}\text{ m}\cdot {\text{s}}^{-1}\right)\left(1.0\times {10}^{-2}\text{ m}\right)\\ =2\times {10}^{-5}\text{ V}\end{array}$

Therefore, the Hall potential difference is $\underset{\_}{2\times {10}^{-5}\text{ V}}$.

(f)

To determine

The direction of current in the circuit in Figure 19.37.

Answer to Problem 33E

The direction of the current is out of the page.

Explanation of Solution

Figure 1 represents the direction of the current in the circuit. Since the drift velocity of electrons is into the page the direction of current will be opposite to that. That is out of the page.

Conclusion:

Therefore, the direction of the current is out of the page.

(g)

To determine

To check whether the Hall potential difference be the same when the carriers are negative.

Answer to Problem 33E

The Hall potential difference will be opposite in sign for a given current direction.

Explanation of Solution

Hall Effect is a conduction phenomenon which is different for different charge carriers. Hall voltage has different polarity for positive and negative charge carriers.  For a given current direction if the negative carriers are causing the Hall voltage instead of the positive carriers, the hall voltage will be opposite in sign.

Conclusion:

Therefore, the Hall potential difference will be opposite in sign for a given current direction.