Chapter 24, Problem 102A

a.

To determine

The magnetic field of the current carrying wire at $0.5\text{ m}$ and comparing it with Earth’s magneticfield.

Answer to Problem 102A

The magnetic field by the wire is $B=4\times {10}^{-6}\text{ T}$ .

Explanation of Solution

Given:

The magnetic field produced by a wire, $B=\frac{\left(2\times {10}^{-7}\text{ T}\cdot \text{m/A}\right)I}{d}$

Current, $I=10\text{ A}$

Distance from the wire, $d=0.5\text{}\text{ m}$

Formula Used:

  $B=\frac{\left(2\times {10}^{-7}\text{ T}\cdot \text{m/A}\right)I}{d}$

I is the current

d is the distance.

Calculations:

  $\text{B=}\frac{\left(2\times {10}^{-7}\text{ T}\text{.m/A}\right)I}{d}$

Putting in values

  $B=\frac{\left(2\times {10}^{-7}\text{ T}\text{.m/A}\right)\left(10\text{ A}\right)}{0.5\text{}\text{ m}}$

  $B=4\times {10}^{-6}\text{ T}$

Earth’s magnetic field is $B=5\times {10}^{-5}\text{ T}$ which is twelve times stronger than the magnetic field by the wire.

Conclusion:

The magnetic field by the wire is $B=4\times {10}^{-6}\text{ T}$ . Earth’s magnetic field is $B=5\times {10}^{-5}\text{ T}$ which is twelve times stronger than the magnetic field by the wire.

b.

To determine

The magnetic field of the power transmission wire at the ground and compare it with magnetic field of the wire at home.

Answer to Problem 102A

The magnetic field by the power line is $B=2\times {10}^{-6}\text{ T}$ .

Explanation of Solution

Given:

The magnetic field by a wire, $B=\frac{\left(2\times {10}^{-7}\text{ T}\cdot \text{m/A}\right)I}{d}$

Current, $I=100\text{ A}$

Distance from the wire, $d=20\text{}\text{ m}$

Formula Used:

  $B=\frac{\left(2\times {10}^{-7}\text{ T}\cdot \text{m/A}\right)I}{d}$

I is the current

d is the distance.

Calculations:

  $B=\frac{\left(2\times {10}^{-7}\text{ T}\cdot \text{m/A}\right)I}{d}$

Substitute values

  $B=\frac{\left(2\times {10}^{-7}\text{ T}\cdot \text{/A}\right)\left(100\text{ A}\right)}{20\text{}\text{ m}}$

  $B=2\times {10}^{-6}\text{ T}$

At home the wire’s magnetic field is $B=4\times {10}^{-6}\text{ T}$ which is twice as stronger as the magnetic field by the power transmission cable.

Conclusion:

The magnetic field by the power line is $B=2\times {10}^{-6}\text{ T}$ . At home the wire’s magnetic field is $B=4\times {10}^{-6}\text{ T}$ which is twice as stronger as the magnetic field by the power transmission cable.