Chapter 19, Problem 67QAP

To determine

(a)

The magnetic field $1.0$ m and 1.0 km from a bolt of lighting can transfer 10 C of charge in $2.0\mu \text{s}$ and also compare the calculated fields with Earth`s magnetic field

Answer to Problem 67QAP

  $\begin{array}{l}\text{Magnetic field at a distance of 1}\text{.0 m = 1}.0\text{ T}\\ \text{Magnetic field at a distance of 1}\text{.0 km}=1.0\times {10}^{-3}\text{ T}\\ \text{Comparison of earth magnetic field with 1}\text{.0 m magnetic field due to bolt =20000}\\ \text{Comparison of earth magnetic field with 1}\text{.0 m magnetic field due to bolt = 20}\end{array}$

Explanation of Solution

Given info:

  $\begin{array}{l}\text{Transferring charge = }10\text{ C}\\ \text{Duration = }2.0\mu \text{s}\end{array}$

Formula used:

  $\begin{array}{l}I=\frac{Q}{t}\\ I=\text{Current}\\ Q=\text{charge}\\ t=\text{time}\end{array}$

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

Calculation:

  $\begin{array}{l}\text{Current produced,}\\ I=\frac{Q}{t}\\ I=\frac{10\text{ C}}{2\times {10}^{-6}\text{ s}}\\ I=5\times {10}^6\text{ A}\end{array}$

  $\begin{array}{l}\text{Magnetic field at a distance of 1}\text{.0 m}\\ B=\frac{{\mu }_{0}I}{2\pi d}\\ B=\frac{4\pi \times {10}^{-7}\times 5\times {10}^6}{2\pi \times 1}\\ B=1.0\text{ T}\end{array}$

  $\begin{array}{l}\text{Magnetic field at a distance of 1}\text{.0 km}\\ B=\frac{{\mu }_{0}I}{2\pi d}\\ B=\frac{4\pi \times {10}^{-7}\times 5\times {10}^6}{2\pi \times 1\times {10}^3}\\ B=1.0\times {10}^{-3}\text{ T}\end{array}$

  $\begin{array}{l}\text{Earth magnetic field = }5.0\times {10}^{-5}\text{ T}\\ \text{ Comparison with 1}\text{.0 m magnetic field due to bolt,}\\ \frac{B}{B_{earth}}=\frac{1\text{ T}}{5.0\times {10}^{-5}\text{ T}}=20000\\ \text{ Comparison with 1}\text{.0 km magnetic field due to bolt,}\\ \frac{B}{B_{earth}}=\frac{1.0\times {10}^{-3}\text{ T}}{5.0\times {10}^{-5}\text{ T}}=20\end{array}$

Conclusion:

Thus, the magnetic field at a distance of 1 m is 1 T and at distance of 1 km $1.0\times {10}^{-3}\text{ T}$ and after comparing of Earth magnetic field with 1.0 m magnetic field due to bolt is 20000 and with 1.0 km is 20

To determine

(b)

Compare the fields in part (a) with the magnetic field produced by a typical household current of 10 A in a very long wire at the same distances from the wire as in (a)

Answer to Problem 67QAP

  $\begin{array}{l}\text{Magnetic field at a distance of 1}\text{.0 m = }2*{10}^{-6}\text{ T}\\ \text{Magnetic field at a distance of 1}\text{.0 km}=2.0*{10}^{-9}\text{ T}\\ \text{Comparison of house hold current 1}\text{.0 m magnetic field to bolt 1}\text{.0 m magnetic field =500000}\\ \text{Comparison of house hold current 1}\text{.0 km magnetic field to bolt 1}\text{.0 km magnetic field =500000}\end{array}$

Explanation of Solution

Given info:

  $\text{Current = }10\text{ A}$

Formula used:

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

Calculation:

  $\begin{array}{l}\text{Magnetic field at a distance of 1}\text{.0 m}\\ B=\frac{{\mu }_{0}I}{2\pi d}\\ B=\frac{4\pi \times {10}^{-7}\times 10}{2\pi \times 1}\\ B=2\times {10}^{-6}\text{ T}\end{array}$

  $\begin{array}{l}\text{Magnetic field at a distance of 1}\text{.0 km}\\ B=\frac{{\mu }_{0}I}{2\pi d}\\ B=\frac{4\pi \times {10}^{-7}\times 10}{2\pi \times 1\times {10}^3}\\ B=2.0\times {10}^{-9}\text{ T}\end{array}$

  $\begin{array}{l}\text{Earth magnetic field = }5.0\times {10}^{-5}\text{ T}\\ \text{ Comparison with 1}\text{.0 m magnetic field due to bolt and 1}\text{.0 m due to household current,}\\ \frac{B_{bolt}}{B_{house}}=\frac{1\text{ T}}{2.0\times {10}^{-6}\text{ T}}=500000\\ \text{Comparison with 1}\text{.0 km magnetic field due to bolt and 1}\text{.0 km due to household current,}\\ \frac{B_{bolt}}{B_{house}}=\frac{1.0\times {10}^{-3}\text{ T}}{2.0\times {10}^{-9}\text{ T}}=500000\end{array}$

Conclusion:

  $\begin{array}{l}\text{Magnetic field at a distance of 1}\text{.0 m = }2\times {10}^{-6}\text{ T}\\ \text{Magnetic field at a distance of 1}\text{.0 km}=2.0\times {10}^{-9}\text{ T}\\ \text{Comparison of house hold current 1}\text{.0 m magnetic field to bolt 1}\text{.0 m magnetic field =500000}\\ \text{Comparison of house hold current 1}\text{.0 km magnetic field to bolt 1}\text{.0 km magnetic field =500000}\end{array}$

To determine

(c)

How close would you have to get to the wire in part (b) for its magnetic field to be the same as the field produced by the lightning bolt at 1.0 km from the bolt

Answer to Problem 67QAP

2 mm

Explanation of Solution

Given info:

  $\text{Current = }10\text{ A}$

Formula used:

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

Calculation:

  $\begin{array}{l}\text{Magnetic field due to bolt at a distance of 1}\text{.0 km = 1}\times {\text{10}}^{\text{-3}}\text{ T}\\ B=\frac{{\mu }_{0}I}{2\pi d}\\ \text{1}\times {\text{10}}^{\text{-3}}\text{ T}=\frac{4\pi \times {10}^{-7}\times 10}{2\pi \times d}\\ d=2\times {10}^{-3}\text{ m}\end{array}$

Conclusion:

Thus, the wire should 2 mm close for its magnetic field to be the same as the field produced by the lightning bolt at 1.0 km from the bolt