Chapter 20: Problem 15
A current of \(20 \mathrm{~A}\) flows through each of two parallel long wires which are \(4 \mathrm{~cm}\) apart. The force exerted per unit length of each wire is (a) \(1 \times 10^{-3} \mathrm{~N}\) (b) \(2 \times 10^{-3} \mathrm{~N}\) (c) \(3 \times 10^{-3} \mathrm{~N}\) (d) \(4 \times 10^{-3} \mathrm{~N}\)
Short Answer
Step by step solution
Identify the Formula
Substitute the Values
Perform the Calculation
Select the Correct Answer
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ampere's Law
\[\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{ ext{enc}}\]where
- \(\mathbf{B}\) is the magnetic field,
- \(d\mathbf{l}\) is a differential element of the loop,
- \(\mu_0\) is the permeability of free space,
- and \(I_{\text{enc}}\) is the current enclosed.
Permeability of Free Space
In practical terms:
- The permeability of free space helps to quantify the magnetic force between current-carrying wires.
- It plays a crucial role in calculating the magnetic field produced by electric currents.
- Understanding this concept is key to grasping how magnetic fields operate in basic electromagnetic systems.
Current and Magnetic Fields
Some important aspects to remember include:
- The right-hand rule helps determine the direction of the magnetic field: if you point your thumb in the direction of current, your curled fingers indicate the magnetic field direction.
- The strength of the magnetic field is proportional to the current flowing through the wire.
- Parallel currents attract each other, while opposite currents repel.
Force per Unit Length
\[F/L = \frac{\mu_0 I_1 I_2}{2\pi d}\]where
- \(F\) is the force,
- \(L\) is the length,
- \(I_1\) and \(I_2\) are the currents in the wires,
- \(d\) is the separation between the wires,
- and \(\mu_0\) is the permeability of free space.
Understanding this helps in calculating how parallel wires carrying current influence each other through magnetic forces.