Q59P :In Figure 20.121 a bar 11 cm lo... [FREE SOLUTION]

Chapter 20: Q59P (page 861)

: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 $7 脳 10^{23} / m^{3}$ , and their mobility is $3 脳 10^{鈭� 5} 鈥� (m/s) / (V/m)$ .
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

Expert verified

$(i) I = 0.0219 A (i i) V_{H} = - 1.76 脳 10^{- 5} V o l t$

Step by step solution

Given Data

$\begin{array}{l} 尝别苍驳迟丑,鈥� L = 11 鈥刢尘 = 11 脳 10^{鈭� 2} 鈥刴 \\ Height = 3 鈥刢尘=3 脳 10^{鈭� 2} 鈥刴 \\ Depth = 2 鈥刢尘=2 脳 10^{鈭� 2} 鈥刴 \\ Voltage V = 1.2 鈥刅 \\ 狈耻尘产别谤鈥刼蹿鈥别濒别肠迟谤辞苍蝉, n = 7 脳 10^{23} {/m}^{3} \\ Mobility, 渭 = 3 脳 10^{鈭� 5} (m/s) / (V/m) \\ B = 1.8 鈥凾 \\ e = 1.6 脳 10^{鈭� 19} C \end{array}$

Concept

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

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

Area,

$\begin{matrix} A = (3 脳 10^{鈭� 2}) (2 脳 10^{鈭� 2}) m^{2} \\ = 6 脳 10^{鈭� 4} m^{2} \end{matrix}$

The following expressions are used to find I,

$\begin{array}{l} I = J 鈰� A \\ J = 蟽鈰� E \\ 蟽 = n e 渭 \\ E = \frac{v}{L} \end{array}$

Substitute the above expressions in I,

$I = n e 渭脳 \frac{v}{L} 脳 A$

$\begin{matrix} I = n e 渭脳 \frac{v}{L} 脳 A \\ = 7 脳 10^{23} {/m}^{3} 脳 1.6 脳 10^{鈭� 19} C 脳 3 脳 10^{鈭� 5} (m/s) / (V/m) 脳 \frac{1.2 鈥刅}{11 脳 10^{鈭� 2} 鈥刴} 脳 6 脳 10^{鈭� 4} m^{2} \\ = 0.0219 鈥凙 \end{matrix}$

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

Hence, the ammeter reading is $滨=0.0219鈥凙$

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

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

$\begin{array}{l} q E = B q V \\ E = B V \\ \frac{V_{H}}{d} = B V \\ V_{H} = B V d \end{array} \begin{matrix} V = \frac{J}{n e} \\ = \frac{n e 渭 E'}{n e} \\ = 渭 \frac{V}{L} \end{matrix}$

To find the voltmeter reading,

$\begin{matrix} V_{H} = B 渭 \frac{V}{L} d \\ = \frac{1.8 脳 3 脳 10^{鈭� 5} 脳 1.2 脳 3 脳 10^{鈭� 2}}{11 脳 10^{鈭� 2}} \\ = 1.76 脳 10^{鈭� 5} 鈥刅辞濒迟 \end{matrix}$

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

Hence, the reading of the voltmeter is $V_{H} = 鈭� 1.76 脳 10^{鈭� 5} 鈥刅辞濒迟$