Point \(P\) is midway between two long, straight, parallel wires that run north- south in a horizontal plane. The distance between the wires is $1.0 \mathrm{cm} .\( Each wire carrics a current of \)1.0 \mathrm{A}$ toward the north. Find the magnitude and direction of the magnetic field at point \(P\)

Short Answer

Expert verified
Answer: The magnitude of the magnetic field at point P is approximately \(2.83 \times 10^{-5} \text{T}\), and the direction is upward at a 45-degree angle with respect to a horizontal line going through point P.

Step by step solution

01

Identify the distance between point P and each wire

Since point P is midway between the two wires, the distance between point P and each wire is equal to half the distance between the wires. The distance between the wires is given as 1.0 cm, so the distance between point P and each wire is 0.5 cm or \(0.005 \text{m}\) (in SI units).
02

Calculate the magnetic field at point P due to each wire using the Biot-Savart law

The magnetic field produced by a long, straight wire carrying a current I at a distance d is given by the Biot-Savart law: \(B = \frac{\mu_0 I}{2 \pi d}\) where \(B\) is the magnetic field, \(\mu_0\) is the permeability of free space (\(4 \pi \times 10^{-7} \text{T}\cdot\text{m/A}\)), \(I\) is the current in the wire, and \(d\) is the distance from the wire to the point where we're calculating the magnetic field. We know that each wire carries a current of 1.0 A and the distance from each wire to point P is 0.005 m. Therefore, the magnetic field at point P due to each wire is: \(B_{\text{each}} = \frac{4 \pi \times 10^{-7} \text{T}\cdot \text{m/A} \times 1.0 \text{A}}{2 \pi \times 0.005 \text{m}} = 2 \times 10^{-5} \text{T}\)
03

Find the total magnetic field at point P by adding the magnetic fields due to each wire

Total magnetic field at point P can be found by adding the magnetic fields produced by both wires. However, since these magnetic fields are not in the same direction (they are perpendicular to each other), they cannot be directly added. Instead, we need to find the vector sum of the magnetic fields. Let's denote the magnetic field produced by the left wire as \(B_{\text{left}}\) and the magnetic field produced by the right wire as \(B_{\text{right}}\). The magnetic field produced by the left wire points into the page, while the magnetic field produced by the right wire points out of the page. To find the total magnetic field at point P, we can compute the vector sum of these magnetic fields, which will be at a 90-degree angle: \(B_{\text{total}} = \sqrt{B_{\text{left}}^2 + B_{\text{right}}^2}\) Since the magnitudes of both magnetic fields are equal, \(B_{\text{left}} = B_{\text{right}} = B_{\text{each}} = 2 \times 10^{-5} \text{T}\), plugging the values into the above formula: \(B_{\text{total}} = \sqrt{(2 \times 10^{-5} \text{T})^2 + (2 \times 10^{-5} \text{T})^2} = 2\sqrt{2} \times 10^{-5} \text{T} \approx 2.83 \times 10^{-5} \text{T}\)
04

Determine the direction of the magnetic field at point P

The magnetic field at point P is the vector sum of the magnetic field due to each wire. Since both magnetic fields are equal and perpendicular to each other, they form a 45-degree angle with the horizontal line connecting the wires. In other words, the magnetic field at point P is going upward at a 45-degree angle with respect to a horizontal line going through point P. In conclusion, the magnitude of the magnetic field at point P is approximately \(2.83 \times 10^{-5} \text{T}\), and the direction is upward at a 45-degree angle with respect to a horizontal line going through point P.

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Most popular questions from this chapter

In an electric motor, a coil with 100 turns of radius \(2.0 \mathrm{cm}\) can rotate between the poles of a magnet. The magnetic field strength is $0.20 \mathrm{T}\(. When the current through the coil is \)50.0 \mathrm{mA},$ what is the maximum torque that the motor can deliver?
An electromagnet is made by inserting a soft iron core into a solenoid. The solenoid has 1800 turns, radius \(2.0 \mathrm{cm},\) and length $15 \mathrm{cm} .\( When \)2.0 \mathrm{A}$ of current flows through the solenoid, the magnetic field inside the iron core has magnitude 0.42 T. What is the relative permeability \(\overline{K_{\mathrm{B}}}\) of the iron core? (See Section 19.10 for the definition of \(\boldsymbol{K}_{\mathbf{B}} .\) )
The strength of Earth's magnetic field, as measured on the surface, is approximately \(6.0 \times 10^{-5} \mathrm{T}\) at the poles and $3.0 \times 10^{-5} \mathrm{T}$ at the equator. Suppose an alien from outer space were at the North Pole with a single loop of wire of the same circumference as his space helmet. The diameter of his helmet is \(20.0 \mathrm{cm} .\) The space invader wishes to cancel Earth's magnetic field at his location. (a) What is the current required to produce a magnetic ficld (due to the current alone) at the center of his loop of the same size as that of Earth's field at the North Pole? (b) In what direction does the current circulate in the loop, CW or CCW, as viewed from above, if it is to cancel Earth's field?
Two long straight wires carry the same amount of current in the directions indicated. The wires cross each other in the plane of the paper. Rank points \(A, B\) \(C,\) and \(D\) in order of decreasing field strength.
A long straight wire carries a current of 3.2 \(A\) in the positive \(x\)-direction. An electron, traveling at \(6.8 × 10 6\) m/s in the positive \(x\) -direction, is 4.6 cm from the wire. What force acts on the electron?
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