Question

:In Figure 20.121 a bar 11 cm long with a rectangular cross section 3 cm high and 2 cm deep is connected to a 1.2 V battery and an ammeter. The resistance of the copper connecting wires and the ammeter, and the internal resistance of the battery, are all negligible compared to the resistance of the bar.

Using large coils not shown on the diagram, a uniform magnetic field of 1.8 T was applied perpendicular to the bar (out of the page, as shown). A voltmeter was connected across the bar, with the connections across the bar carefully placed directly across from each other.

The mobile charges in the bar have charge +e, their density is $\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{/}{\mathbf{\text{m}}}^{\mathbf{\text{3}}}$, and their mobility is $\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{\left(}\mathbf{\text{m/s}}\mathbf{\right)}\mathbf{/}\mathbf{\left(}\mathbf{\text{V/m}}\mathbf{\right)}$.

Predict the readings of the voltmeter and ammeter, including signs. Explain carefully, using diagrams to support your explanation. Remember that a voltmeter reads positive if the + terminal is connected to higher potential, and that an ammeter reads positive if conventional current enters the + terminal.

Short answer

$$\begin{gathered} \left(i\right)I=0.0219A \\ \left(ii\right)V_H=-1.76\times {10}^{-5}Volt \end{gathered}$$

Step-by-step solution

Given Data

$\begin{array}{l}\text{Length,\hspace{0.33em}}L=11\text{\hspace{0.33em}cm}=11\times {10}^{-2}\text{\hspace{0.33em}m}\\ \text{Height}=3\text{\hspace{0.33em}cm=3}\times {10}^{-2}\text{\hspace{0.33em}m}\\ \text{Depth}=2\text{\hspace{0.33em}cm=2}\times {10}^{-2}\text{\hspace{0.33em}m}\\ \text{Voltage}V=1.2\text{\hspace{0.33em}V}\\ \text{Number\hspace{0.33em}of\hspace{0.33em}electrons},n=7\times {10}^{23}{\text{/m}}^{\text{3}}\\ \text{Mobility},\mu =3\times {10}^{-5}\left(\text{m/s}\right)/\left(\text{V/m}\right)\\ B=1.8\text{\hspace{0.33em}T}\\ e=1.6\times {10}^{-19}\text{C}\end{array}$

Concept

Ammeter measures the current flow. Voltmeter measures voltage in electric circuit.

Step 3(i): Predict the readings of the ammeter

Area,

$\begin{array}{c}A=\left(\text{3}\times {10}^{-2}\right)\left(\text{2}\times {10}^{-2}\right){\text{m}}^{\text{2}}\\ =6\times {10}^{-4}{\text{m}}^{\text{2}}\end{array}$

The following expressions are used to find I,

$\begin{array}{l}I=J\cdot A\\ J=\sigma \cdot E\\ \sigma =ne\mu \\ E=\frac{v}{L}\end{array}$

Substitute the above expressions in I,

$I=ne\mu \times \frac{v}{L}\times A$

$\begin{array}{c}I=ne\mu \times \frac{v}{L}\times A\\ =7\times {10}^{23}{\text{/m}}^{\text{3}}\times 1.6\times {10}^{-19}\text{C}\times 3\times {10}^{-5}\left(\text{m/s}\right)/\left(\text{V/m}\right)\times \frac{1.2\text{\hspace{0.33em}V}}{11\times {10}^{-2}\text{\hspace{0.33em}m}}\times 6\times {10}^{-4}{\text{m}}^{\text{2}}\\ =0.0219\text{\hspace{0.33em}A}\end{array}$

As the +ve terminal of the ammeter is connected to the +ve terminal of the battery and –ve terminal of the battery is connected to the –ve terminal of the ammeter. So the reading of ammeter is +ve.

Hence, the ammeter reading is$\text{I=0.0219\hspace{0.33em}A}$

Step 4(ii): Predict the readings of the voltmeter

The following expressions are used to find the reading of the voltmeter,

$$\begin{gathered} \begin{array}{l}qE=BqV\\ E=BV\\ \frac{V_H}{d}=BV\\ V_H=BVd\end{array} \\ \begin{array}{c}V=\frac{J}{ne}\\ =\frac{ne\mu E\text{'}}{ne}\\ =\mu \frac{V}{L}\end{array} \end{gathered}$$

To find the voltmeter reading,

$\begin{array}{c}{V}_H=B\mu \frac{V}{L}d\\ =\frac{1.8\times 3\times {10}^{-5}\times 1.2\times \text{3}\times {10}^{-2}}{11\times {10}^{-2}}\\ =1.76\times {10}^{-5}\text{\hspace{0.33em}Volt}\end{array}$

But the connection of the voltmeter is reversed in this case. The –ve terminal is connected with the +ve surface and +ve terminal is connected with –ve surface of the bar. So the reading of voltmeter is –ve.

Hence, the reading of the voltmeter is $V_H=-1.76\times {10}^{-5}\text{\hspace{0.33em}Volt}$