Chapter 20: Problem 73
\(\bullet\) Three long, straight electrical cables, running north and south, are tightly enclosed in an insulating sheath. One of the cables carries a 23.0 A current southward; the other two carry currents of 17.5 \(\mathrm{A}\) and 11.3 A northward. Use Ampere's law to calculate the magnitude of the magnetic field at a distance of 10.0 \(\mathrm{m}\) from the cables.
Short Answer
Step by step solution
Understand the context
List given values
Calculate the net current using direction
Apply Ampere's Law
Compute the magnetic field, B
Finalize the magnitude of B
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Magnetic Field Calculation
Ampere's Law states that the magnetic field around a closed loop is proportional to the electric current passing through the loop. Mathematically, this is \[ B = \frac{\mu_0 \cdot I_{net}}{2 \pi \cdot r} \]where \( \mu_0 \) is the permeability of free space (\( 4 \pi \times 10^{-7} \, T \cdot m/A \)), \( I_{net} \) is the net current, and \( r \) is the distance from the wire.
- Start by calculating the net current considering the direction of each current.
- Apply the condition of Ampere's Law in a loop passing through the cables.
- Solve for the magnetic field at the point of interest.
Currents in Electrical Cables
When calculating effects like the magnetic field, the direction of current is crucial. For currents:
- Southward currents (opposite our reference direction) are taken as negative.
- Northward currents (along our reference direction) are positive.
Net Current
In our scenario, you calculate net current \( I_{net} \) as follows:
- Sum the northward currents (positive values): \( 17.5 \, A + 11.3 \, A = 28.8 \, A \).
- Subtract the southward current (negative value): \( 28.8 \, A - 23.0 \, A = 5.8 \, A \).
Magnetic Field Magnitude
Using the formula \[ B = \frac{\mu_0 \cdot I_{net}}{2 \pi \cdot r} \],substitute:
- \( \mu_0 = 4 \pi \times 10^{-7} \, T \cdot m/A \)
- \( I_{net} = 5.8 \, A \)
- and \( r = 10.0 \, m \)
\[ B = \frac{4 \pi \times 10^{-7} \, T \cdot m/A \times 5.8 \, A}{2 \pi \times 10.0 \, m} \]
The computed magnitude becomes \( 1.16 \times 10^{-7} \, T \). This magnetic field magnitude shows the influence of the currents at the specified distance.