Chapter 24, Problem 103A

To determine

The scale drawing of the magnetic fields by the two wires and find their vector sum. Also find its magnitude and direction

Answer to Problem 103A

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

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, $I=10\text{ A}$

Distance from the wire, $d=0.1\text{ m}$

Formula Used:

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

Iis the current

d is the distance.

Calculations:

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

The magnetic field by each wire at $0.1\text{}\text{ m}$

Putting in values

  $B=\frac{\left(2\times {10}^{-7}\text{ T}\cdot \text{m/A}\right)\left(10\text{ A}\right)}{0.1\text{}\text{ m}}$

  $B=2\times {10}^{-5}\text{ T}$

The horizontal components of the magnetic fields by each wire is $B\sin \theta$ .

  $\sin \theta =\frac{0.005\text{ m}}{0.10\text{ m}}=0.05\text{ m}$

  $\begin{array}{c}B\sin \theta =\left(2\times {10}^{-6}\text{ T}\right)0.05\text{ m}\\ \text{=1}\times {10}^{-6}\text{ T}\end{array}$

Using the right-hand rule, the thumb directs towards the flow of the current, then the curled fingers represent the direction of the magnetic field.

Therefore, using the right hand rule the magnetic field by both the wire is towards the left.

Both the wire contribute equally, hence

Net magnetic field is $B\text{=}\left(2\right)\left(\text{1}\times {10}^{-6}\text{ T}\right)$

  $B=2\times {10}^{-6}\text{ T}$

Conclusion:

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