Chapter 37, Problem 21A

The magnets are moved into the wire coils in identical quick fashion. Voltage induced in each coil causes a current to flow, as indicated on the galvanometer. Neglect the electrical resistance of the loops in the coil.

a. Rank from greatest to least the reading on the galvanometer.

b. Make the same ranking, only this time for each coil having twice as many loops as in part (a).

To determine

To arrange: the reading of galvanometer in decreasing order.

Answer to Problem 21A

The decreasing order of the galvanometer is :

$\mathrm{B}>C>A$

Explanation of Solution

Given:

System A:

  $1magnetand3coilloops$

System B:

  $1magnetand\text{ 5 }coilloops$

System C:

  $\text{2 }magnetand\text{ 2 }coilloops$

Concept Used:

The relation between the voltage and the number of coils is given as :

$\begin{array}{l}V=NA\frac{\Delta B}{\Delta t}\\ N=numberofcoilloops\\ A=areaofcrosssection\\ \Delta B=changeinmagneticfield\\ \Delta t=changeintime\end{array}$

Calculation:

For system A:

  $\begin{array}{l}N=3\\ A=A\end{array}$

The voltage produced is calculated as:

${\mathrm{V}}_A=3A\frac{\Delta \mathrm{B}}{\Delta \mathrm{t}}\mathrm{................................................................}(1)$

For system B:

$\begin{array}{l}N=5\\ A=A\end{array}$

The voltage produced is calculated as:

${\mathrm{V}}_B=5A\frac{\Delta \mathrm{B}}{\Delta \mathrm{t}}\mathrm{................................................................}(2)$

For system C:

$\begin{array}{l}N=2\\ A=2A\end{array}$

The voltage produced is calculated as:

${\mathrm{V}}_C=4A\frac{\Delta \mathrm{B}}{\Delta \mathrm{t}}\mathrm{................................................................}(3)$

From (1) , (2) & (3) we get:

$\begin{array}{l}{V}_B>V_C>V_A\\ as\\ V\propto I(Resistanceneglected)\\ I_B>I_C>I_A\end{array}$

Conclusion:

The galvanometer readings greatest to least:

$\mathrm{B}>C>A$

To determine

To arrange: the reading of the galvanometer in decreasing order when the loops having twice coils in all systems.

Answer to Problem 21A

The decreasing order of the galvanometer is :

$\mathrm{B}>C>A$

Explanation of Solution

Given:

System A:

  $1magnetand\text{ 6 }coilloops$

System B:

  $1magnetand\text{ 10 }coilloops$

System C:

  $\text{2 }magnetand\text{ 4 }coilloops$

Concept Used:

The relation between the voltage and the number of coils is given as :

$\begin{array}{l}V=NA\frac{\Delta B}{\Delta t}\\ N=numberofcoilloops\\ A=areaofcrosssection\\ \Delta B=changeinmagneticfield\\ \Delta t=changeintime\end{array}$

Calculation:

For system A:

$\begin{array}{l}N=6\\ A=A\end{array}$

The voltage produced is calculated as:

${\mathrm{V}}_{A\text{'}}=6A\frac{\Delta \mathrm{B}}{\Delta \mathrm{t}}\mathrm{................................................................}(4)$

For system B:

$\begin{array}{l}N=10\\ A=A\end{array}$

The voltage produced is calculated as:

${\mathrm{V}}_{B\text{'}}=10A\frac{\Delta \mathrm{B}}{\Delta \mathrm{t}}\mathrm{................................................................}(5)$

For system C:

$\begin{array}{l}N=4\\ A=2A\end{array}$

The voltage produced is calculated as:

${\mathrm{V}}_C=8A\frac{\Delta \mathrm{B}}{\Delta \mathrm{t}}\mathrm{................................................................}(6)$

From (4) , (5) & (6) we get:

$\begin{array}{l}{V}_{B\text{'}}>V_{C\text{'}}>V_{A\text{'}}\\ as\\ V\propto I(Resistanceneglected)\\ I_{B\text{'}}>I_{C\text{'}}>I_{A\text{'}}\end{array}$

Conclusion:

The galvanometer readings greatest to least:

$\mathrm{B}\text{'}>C\text{'}>A\text{'}$