Chapter 19, Problem 78QAP

To determine

(a)

The magnitude of the magnetic field the cord produces midway between the two wires

Answer to Problem 78QAP

Magnitude of the magnetic field the cord produces midway between the two wires= $1.36\times {10}^{-4}\text{ T}$

Explanation of Solution

Given info:

  $\begin{array}{l}\text{Voltage = }110\text{ V}\\ \text{The electric power of the lamp = }75\text{ W}\\ \text{two insulated parallel wires 4}.0\text{ mm apart}\\ \text{Earth's magnetic field = 5}\times {\text{10}}^{\text{-5}}\text{ T}\end{array}$

Formula used:

  $\begin{array}{l}P=VI\\ V=\text{potential}\\ P=\text{Power}\\ I=\text{current}\end{array}$

  $\begin{array}{l}B=\frac{{\mu }_{0}I}{2\pi d}\\ B=\text{Magnetic field}\\ {\mu }_0=\text{permeability of free space}\\ I=\text{Current in wire}\\ d=\text{Radius from the wire}\end{array}$

Calculation:

  $\begin{array}{l}P=VI\\ 75=110\times I\\ I=0.68\text{ A}\end{array}$

  $\begin{array}{l}B=\frac{{\mu }_{0}I}{2\pi d}\\ B=\frac{4\pi \times {10}^{-7}\times 0.68}{2\pi \times 2\times {10}^{-3}}\\ B=0.68\times {10}^{-4}\text{ T}\\ \text{Both B due to two wires are equal in magnitude and to same direction}\\ B_{Total}=2\times 0.68\times {10}^{-4}\text{ T}\\ B_{Total}=1.36\times {10}^{-4}\text{ T}\end{array}$

Conclusion:

Magnitude of the magnetic field the cord produces midway between the two wires= $1.36\times {10}^{-4}\text{ T}$

To determine

(b)

The magnitude of the magnetic field the cord produces 2.0 mm from one of the wires in the same plane in which the two wires lie

Answer to Problem 78QAP

Magnitude of the magnetic field the cord produces 2.0 mm from one of the wires in the same plane in which the two wires lie= $0.45*{10}^{-4}\text{ T}$

Explanation of Solution

Given info:

  $\begin{array}{l}\text{Voltage = }110\text{ V}\\ \text{The electric power of the lamp = }75\text{ W}\\ \text{two insulated parallel wires 4}.0\text{ mm apart}\\ {\text{Earth's magnetic field = 5*10}}^{\text{-5}}\text{ T}\end{array}$

Formula used:

  $\begin{array}{l}P=VI\\ V=\text{potential}\\ P=\text{Power}\\ I=\text{current}\end{array}$

  $\begin{array}{l}B=\frac{{\mu }_{0}I}{2\pi d}\\ B=\text{Magnetic field}\\ {\mu }_0=\text{permeability of free space}\\ I=\text{Current in wire}\\ d=\text{Radius from the wire}\end{array}$

Calculation:

For the wire on the right side d =2 mm

  $\begin{array}{l}B=\frac{{\mu }_{0}I}{2\pi d}\\ B=\frac{4\pi \times {10}^{-7}\times 0.68}{2\pi \times 2\times {10}^{-3}}\\ B=0.68\times {10}^{-4}\text{ T}\end{array}$

Field is directed outwards.

For the wire on the left side d =2+4 mm=6 mm

  $\begin{array}{l}B=\frac{{\mu }_{0}I}{2\pi d}\\ B=\frac{4\pi *{10}^{-7}*0.68}{2\pi *6*{10}^{-3}}\\ B=0.23*{10}^{-4}\text{ T}\end{array}$

Field is directed inwards.

  $B_{total}=0.68*{10}^{-4}\text{ T}-0.23*{10}^{-4}\text{ T=}0.45*{10}^{-4}\text{ T}$

Field is directed outwards.

Conclusion:

Magnitude of the magnetic field the cord produces 2.0 mm from one of the wires in the same plane in which the two wires lie= $0.45*{10}^{-4}\text{ T}$

To determine

(c)

Comparison of previous magnetic fields with earth's magnetic field

Answer to Problem 78QAP

For case a),

  $\frac{B}{B_{earth}}=2.72$

For case b),

  $\frac{B}{B_{earth}}=0.9$

Explanation of Solution

Given info:

  $\begin{array}{l}\text{Voltage = }110\text{ V}\\ \text{The electric power of the lamp = }75\text{ W}\\ \text{two insulated parallel wires 4}.0\text{ mm apart}\\ {\text{Earth's magnetic field = 5*10}}^{\text{-5}}\text{ T}\end{array}$

Formula used:

  $\begin{array}{l}P=VI\\ V=\text{potential}\\ P=\text{Power}\\ I=\text{current}\end{array}$

  $\begin{array}{l}B=\frac{{\mu }_{0}I}{2\pi d}\\ B=\text{Magnetic field}\\ {\mu }_0=\text{permeability of free space}\\ I=\text{Current in wire}\\ d=\text{Radius from the wire}\end{array}$

Calculation:

For case a),

  $\frac{B}{B_{earth}}=\frac{1.36*{10}^{-4}\text{ T}}{5*{10}^{-5}\text{ T}}=2.72$

For case b),

  $\frac{B}{B_{earth}}=\frac{0.45*{10}^{-4}\text{ T}}{5*{10}^{-5}\text{ T}}=0.9$

Conclusion:

For case a),

  $\frac{B}{B_{earth}}=2.72$

For case b),

  $\frac{B}{B_{earth}}=0.9$

To determine

(d)

Magnetic force do the two wires exert on one another

Answer to Problem 78QAP

Force= $0.46*{10}^{-4}\text{ N}$

Force is very small

Explanation of Solution

Given info:

  $\begin{array}{l}\text{Voltage = }110\text{ V}\\ \text{The electric power of the lamp = }75\text{ W}\\ \text{two insulated parallel wires 4}.0\text{ mm apart}\\ {\text{Earth's magnetic field = 5*10}}^{\text{-5}}\text{ T}\end{array}$

Formula used:

  $\begin{array}{l}F=\frac{{\mu }_0{I}^{2}l}{2\pi d}\\ B=\text{Magnetic field}\\ {\mu }_0=\text{permeability of free space}\\ I=\text{Current in wire}\\ d=\text{Radius from the wire}\\ l=\text{length of wire}\end{array}$

Calculation:

Force on each wire.

  $\begin{array}{l}F=\frac{{\mu }_0{I}^{2}l}{2\pi d}\\ F=\frac{4\pi *{10}^{-7}*{0.68}^2*2}{2\pi *4*{10}^{-3}}\\ F=0.46*{10}^{-4}\text{ N}\end{array}$

Force is very small

Conclusion:

Force= $0.46*{10}^{-4}\text{ N}$

Force is very small