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Two long, straight parallel wires are separated by a distance of \(20.0 \mathrm{~cm}\). Each wire carries a current of \(10.0 \mathrm{~A}\) in the same direction. What is the magnitude of the resulting magnetic field at a point that is \(12.0 \mathrm{~cm}\) from each wire?

Short Answer

Expert verified
Answer: The total magnetic field at the point is 3.34 × 10⁻⁴ T.

Step by step solution

01

Identify the given values

In this exercise, we are given the current \(I\) carried by both wires as \(10.0 \mathrm{~A}\), the distance between the wires \(d = 20.0 \mathrm{~cm}\), and the distance from each wire to the point where the magnetic field is being measured, \(r = 12.0 \mathrm{~cm}\).
02

Convert the distances to meters

To use the formula for the magnetic field, we need to convert the given distances from centimeters to meters: \(d = 0.20 \mathrm{~m}\) and \(r = 0.12 \mathrm{~m}\).
03

Calculate the magnetic field created by each wire

Using the formula for the magnetic field created by a long, straight current-carrying wire, we can calculate the magnetic field created by each wire at the given point: $$B = \frac{\mu_0I}{2\pi r} = \frac{4\pi × 10^{-7} Tm/A × 10.0 A}{2\pi × 0.12 m} = \frac{4\pi × 10^{-6} T}{0.24\pi}$$ Canceling the \(\pi\) gives: $$B = \frac{10^{-5} T}{0.06}$$ Now calculating the result: $$B = 1.67 × 10^{-4} \mathrm{T}$$ Thus, each wire creates a magnetic field of \(1.67 × 10^{-4} \mathrm{T}\) at the given point.
04

Determine the vector direction of the magnetic fields

Since both wires carry current in the same direction and are parallel to each other, the magnetic fields produced by each wire at the given point will also be in the same direction. This means that the total magnetic field at the point can be found by simply adding the magnitudes of the magnetic fields created by each wire.
05

Calculate the total magnetic field at the given point

Now that we know the magnetic field created by each wire at the given point, we can find the total magnetic field by adding the magnitudes of the magnetic fields: $$B_\text{total} = B_1 + B_2 = 1.67 \times 10^{-4} T + 1.67 \times 10^{-4} T = 3.34 \times 10^{-4} \mathrm{T}$$ The total magnetic field at the point that is \(12.0 \mathrm{~cm}\) from each wire is \(\mathbf{3.34 × 10^{-4} T}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Field Calculation
Calculating magnetic fields involves determining the influence of moving charges, typically in currents, on the space around them. Magnetic field strength can be crucial when assessing the impact of these forces on nearby objects or environments. To compute this accurately, essential parameters such as the current magnitude, the distance from the current, and the medium in which the field is present must be known. Each ampere of current contributes to the magnetic field. Additionally, the closer a point is to the current, the stronger the field is at that point.

For straight, long wires carrying a current, the magnetic field,\[ B = \frac{\mu_0 I}{2 \pi r} \] where \( \mu_0 \) is the permeability of free space, \( I \) is the current, and \( r \) is the distance from the wire. This equation helps in determining the strength of the field at any specific point near a current-carrying wire.
Biot-Savart Law
The Biot-Savart Law is fundamental in electromagnetism, providing the method to calculate the magnetic field generated by any given steady current. It emphasizes the contributions of small segments of current-carrying wire to the overall magnetic field. The law states that the magnetic field \( B \) at a point in space is directly proportional to the current \( I \), the length of the wire segment, and inversely proportional to the square of the distance from the segment to the point.

Mathematically, it's described as: \[ dB = \frac{\mu_0}{4\pi} \frac{Id\vec{l} \times \hat{r}}{r^2} \]This equation is integral to calculating magnetic fields in more complex circuits beyond straightforward straight-line currents. Understanding this helps in visualizing how segments of current influence a point in space, particularly in cases where wire configurations create more varied field dynamics.
Parallel Currents
When two parallel wires carry current in the same direction, they exert attractive forces on each other due to their magnetic fields. This principle not only applies to current in wires but is a fascinating instance of electromagnetism showing how moving charges create magnetic fields that can interact with other moving charges.

If two wires are carrying current, their magnetic fields will either attract or repel each other. With currents flowing in the same direction, the result is an attractive force between the wires. Conversely, currents in opposite directions result in repulsion. For anyone studying physics, it's crucial to remember that this attraction or repulsion operates through the generated magnetic fields, not directly between the electrons themselves.
Vector Addition in Physics
In the realm of physics, vectors are essential for combining forces, velocities, or fields. When multiple effects come together, their cumulative impact can be determined through vector addition. This process involves considering both magnitude and direction, ensuring a comprehensive calculation.

For magnetic fields, like in our exercise with parallel wires, the fields from each wire are vectors. Because the wires carry currents in the same direction, their magnetic fields at a point add together. You achieve this by adding their magnitudes given they're oriented similarly. Simply put, when dealing with vectors, orientation is just as critical as size. Always ensure you're considering both aspects to accurately determine the total effect when vectors are involved.

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Most popular questions from this chapter

A long, straight cylindrical tube of inner radius \(a\) and outer radius \(b\), carries a total current \(i\) uniformly across its cross section. Determine the magnitude of the magnetic field from the tube at the midpoint between the inner and outer radii.

A hairpin configuration is formed of two semiinfinite straight wires that are \(2.00 \mathrm{~cm}\) apart and joined by a semicircular piece of wire (whose radius must be \(1.00 \mathrm{~cm}\) and whose center is at the origin of \(x y z\) -coordinates). The top straight wire is along the line \(y=1.00 \mathrm{~cm},\) and the bottom straight wire is along the line \(y=-1.00 \mathrm{~cm} ;\) these two wires are in the left side \((x<0)\) of the \(x y\) -plane. The current in the hairpin is \(3.00 \mathrm{~A},\) and it is directed toward the right in the top wire, clockwise around the semicircle, and to the left in the bottom wire. Find the magnetic field at the origin of the coordinate system.

Two long, parallel wires separated by a distance, \(d\), carry currents in opposite directions. If the left-hand wire carries a current \(i / 2,\) and the right-hand wire carries a current \(i\), determine where the magnetic field is zero.

A toy airplane of mass \(0.175 \mathrm{~kg}\), with a charge of \(36 \mathrm{mC}\), is flying at a speed of \(2.8 \mathrm{~m} / \mathrm{s}\) at a height of \(17.2 \mathrm{~cm}\) above and parallel to a wire, which is carrying a 25 - A current; the airplane experiences some acceleration. Determine this acceleration.

A long, straight wire carries a current of 2.5 A. a) What is the strength of the magnetic field at a distance of \(3.9 \mathrm{~cm}\) from the wire? b) If the wire still carries \(2.5 \mathrm{~A}\), but is used to form a long solenoid with 32 turns per centimeter and a radius of \(3.9 \mathrm{~cm}\) what is the strength of the magnetic field at the center of the solenoid?

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