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Two long straight wires are perpendicular to the page and separated by distance \(d_{1}=0.75\) \(\mathrm{cm}\). Wire 1 carries \(6.5 \mathrm{~A}\) into the page. What are the (a) magnitude and (b) direction (into or out of the page) of the current in wire 2 if the net magnetic field due to the two currents is zero at point \(P\) located at distance \(d_{2}=1.50 \mathrm{~cm}\) from wire \(2 ?\)

Short Answer

Expert verified
Current in wire 2 is 4.33 A out of the page.

Step by step solution

01

Calculate Magnetic Field from Wire 1 at Point P

Using the formula for the magnetic field due to a long straight wire, \(B = \frac{\mu_0 I}{2 \pi r}\), we first calculate the magnetic field at point P due to wire 1. The current in wire 1, \(I_1 = 6.5\) A, and the distance from wire 1 to point P is \((d_1 + d_2)\), which is \((0.75\, \text{cm} + 1.50\, \text{cm}) = 2.25\, \text{cm} = 0.0225\, \text{m}\). Thus, the magnetic field is \(B_1 = \frac{(4\pi \times 10^{-7}) \times 6.5}{2 \pi \times 0.0225}\).
02

Set Net Magnetic Field to Zero

For the net magnetic field at point P to be zero, the magnetic field due to wire 2, \(B_2\), must be equal in magnitude but opposite in direction to \(B_1\). Thus, \(B_2 = -B_1\). We already have \(B_1\) calculated from Step 1, and we know the formula for \(B_2 = \frac{\mu_0 I_2}{2\pi d_2}\). Hence, \(\frac{\mu_0 I_2}{2\pi \times 0.015} = \frac{\mu_0 \times 6.5}{2\pi \times 0.0225}\).
03

Solve for Current in Wire 2 (I_2)

Set the equations from Step 2 equal to each other: \(\frac{I_2}{0.015} = \frac{6.5}{0.0225}\). Solving for \(I_2\), we multiply both sides by \(0.015\):\[ I_2 = \frac{6.5 \times 0.015}{0.0225} \]. Simplifying, \(I_2 ≈ 4.33\) A.
04

Determine the Direction of Current in Wire 2

Using the right-hand rule, we determine the direction of the current in wire 2. Wire 1’s magnetic field has to be canceled out, thus wire 2 must have current flowing in the direction that produces a magnetic field opposite to that of wire 1 at point P. If wire 1's field points, say, into the page at P, wire 2's field must consequently point out of the page. As wire 1's field would be clockwise, the current in wire 2 is counterclockwise, hence out of the page.

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Key Concepts

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

Current in a Wire
Understanding the concept of current in a wire is crucial when studying magnetic fields. An electric current is essentially the flow of electric charge through a conductor. In the context of wires, we usually speak about the
  • amount of charge flowing
  • how fast it is moving
which is measured in amperes (A). For example, wire 1 in our exercise carries a current of 6.5 A. This means a significant flow of electric charges is moving through the wire each second.

In a long, straight wire, this moving charge creates a magnetic field around it. This magnetic field spins around the wire, resembling concentric circles. The strength of this field can be calculated using the formula: \[ B = \frac{\mu_0 I}{2 \pi r} \]where:
  • \(B\) is the magnetic field
  • \(\mu_0\) is the permeability of free space
  • \(I\) is the current in amperes
  • \(r\) is the distance from the wire to the point where the magnetic field is measured
Right-Hand Rule
The right-hand rule is a simple yet powerful tool that helps us visualize the direction of the magnetic field created by a current. This rule states that if you curl the fingers of your right hand in the direction of the current flow in the wire, your thumb will point in the direction of the magnetic field.

For our exercise, we're looking at two wires: one with a current going into the page. If we apply the right-hand rule here,
  • curling our fingers into the page aligns them along the circles of the magnetic field around wire 1.
  • Your thumb will point towards the observed magnetic field's direction.
This rule helps us determine that the field at point P is active and in a direction that depends on the current's flow in the wires. When determining the direction of the magnetic field contributed by wire 2 to counteract wire 1's field, the right hand's orientation will be opposite, considering it needs to balance out.
Superposition of Magnetic Fields
The superposition principle allows us to calculate the net magnetic field at a point by adding together the individual magnetic fields from each current. This is immensely helpful when multiple currents, like in our exercise, influence a point in space.

At point P, we want the net magnetic field to be zero, which involves
  • calculating the magnetic field from wire 1
  • understanding the requirement for wire 2's field to counter wire 1 precisely
With superposition, the idea is that these fields simply add up: their directions and magnitudes directly combined.

For the fields to cancel out at point P, wire 2’s field should be equal in magnitude but opposite in direction to that of wire 1. Using the magnetic field formula for our specific contexts, we set up the situation such that \[ \frac{\mu_0 I_2}{2\pi d_2} + \left(-\frac{\mu_0 \times 6.5}{2\pi \times 0.0225}\right)=0 \]. Using this approach, we solved for the current needed in wire 2. Superposition helps simplify complex electromagnetic interactions by focusing on balance and equivalence.

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Most popular questions from this chapter

A 10 -gauge bare copper wire \((2.6 \mathrm{~mm}\) in diameter \()\) can carry a current of 50 A without overheating. For this current, what is the magnitude of the magnetic field at the surface of the wire?

A long solenoid has 100 turns/cm and carries current \(i\). An electron moves within the solenoid in a circle of radius \(2.30 \mathrm{~cm}\) perpendicular to the solenoid axis. The speed of the electron is \(0.0460 c(c=\) speed of light \() .\) Find the current \(i\) in the solenoid.

Shows wire 1 in cross section; the wire is long and straight, carries a current of \(4.00 \mathrm{~mA}\) out of the page, and is at distance \(d_{1}=2.40 \mathrm{~cm}\) from a surface. Wire 2, which is parallel to wire 1 and also long, is at horizontal distance \(d_{2}=5.00 \mathrm{~cm}\) from wire 1 and carries a current of \(6.80\) \(\mathrm{mA}\) into the page. What is the \(x\) component of the magnetic force per unit length on wire 2 due to wire \(1 ?\)

One long wire lies along an \(x\) axis and carries a current of 30 A in the positive \(x\) direction. A second long wire is perpendicular to the \(x y\) plane, passes through the point \((0,4.0 \mathrm{~m}, 0)\), and carries a current of \(40 \mathrm{~A}\) in the positive \(z\) direction. What is the magnitude of the resulting magnetic field at the point \((0,2.0 \mathrm{~m}, 0) ?\)

Shows a snapshot of a proton moving at velocity \(\vec{v}=(-200 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) toward a long straight wire with current \(i=\) \(350 \mathrm{~m} \mathrm{~A}\). At the instant shown, the proton's distance from the wire is \(d=2.89 \mathrm{~cm} .\) In unit-vector nota- tion, what is the magnetic force on the proton due to the current?

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