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Chapter 28: Problem 77

A horizontally oriented coil of wire of radius \(5.00 \mathrm{~cm}\) and carrying a current, \(i\), is being levitated by the south pole of a vertically oriented bar magnet suspended above the center of the coil. If the magnetic field on all parts of the coil makes an angle \(\theta\) of \(45.0^{\circ}\) with the vertical, determine the magnitude and the direction of the current needed to keep the coil floating in midair. The magnitude of the magnetic field is \(B=0.0100 \mathrm{~T}\), the number of turns in the coil is \(N=10.0\), and the total coil mass is \(10.0 \mathrm{~g}\).

Short Answer

Expert verified
Answer: The magnitude of the current needed is \(6.34 \mathrm{~A}\), and the current should flow in a clockwise direction when viewed from above.

Step by step solution

01

Calculate the Gravitational Force

To calculate the force of gravity acting on the coil, we can use the formula: $$F_g = m \cdot g$$ where \(F_g\) is the gravitational force, \(m\) is the mass of the coil, and \(g\) is the acceleration due to gravity (approximately \(9.81 \mathrm{~m/s^2}\)). Given the mass \(m = 10.0 \mathrm{~g} = 0.010 \mathrm{~kg}\), we can calculate the gravitational force: $$F_g = 0.010 \mathrm{~kg} \times 9.81 \mathrm{~m/s^2} = 9.81 \times 10^{-2} \mathrm{N}$$
02

Calculate the Magnetic Force

The magnetic force acting on the coil can be calculated using the formula: $$F_m = B \cdot I \cdot L \cdot \sin(\theta)$$ where \(F_m\) is the magnetic force, \(B\) is the magnetic field, \(I\) is the current, \(L\) is the length of the wire, and \(\theta\) is the angle between the magnetic field and the vertical direction. In our case, \(B = 0.0100 \mathrm{~T}\), \(\theta = 45.0^{\circ}\), and since there are \(N = 10.0\) turns in the coil, the total length of the wire in the field is $$L = 2\pi r \cdot N = 2\pi \cdot 0.05 \mathrm{~m} \cdot 10.0 = \pi \mathrm{~m}$$
03

Equate Magnetic Force to Gravitational Force

To keep the coil levitated, we need the magnetic force to be equal to the gravitational force: $$F_m = F_g$$ Using the magnetic force formula, we have: $$B \cdot I \cdot L \cdot \sin(\theta) = 9.81 \times 10^{-2} \mathrm{N}$$ Now, we can solve for the current \(I\): $$I = \frac{9.81 \times 10^{-2} \mathrm{N}}{0.0100 \mathrm{~T} \cdot \pi \mathrm{~m} \cdot \sin(45.0^{\circ})}$$
04

Calculate the Current

Now, let's plug in the values and find the magnitude of the current needed: $$I = \frac{9.81 \times 10^{-2} \mathrm{N}}{0.0100 \mathrm{~T} \cdot \pi \mathrm{~m} \cdot \frac{\sqrt{2}}{2}} = \frac{9.81 \times 10^{-2} \mathrm{N} \cdot 2 \sqrt{2}}{0.0100 \mathrm{~T} \cdot \pi \mathrm{~m}}$$ $$I = 6.34 \mathrm{~A}$$
05

Determine the Direction of the Current

To determine the direction of the current, we need to consider the orientation of the magnetic field and the levitating force. Since the south pole of the magnet is above the coil, the magnetic field lines are directed downwards. By using the Right Hand Rule, we can determine that the current needs to flow in a clockwise direction when viewed from above for the magnetic force to act upwards and to counteract the gravitational force. So, to keep the coil floating in midair, the required current is \(6.34 \mathrm{~A}\) flowing in a clockwise direction.

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Most popular questions from this chapter

Two identical coaxial coils of wire of radius \(20.0 \mathrm{~cm}\) are directly on top of each other, separated by a 2.00 -mm gap. The lower coil is on a flat table and has a current \(i\) in the clockwise direction; the upper coil carries an identical current and has a mass of \(0.0500 \mathrm{~kg} .\) Determine the magnitude and the direction that the current in the upper coil has to have to keep the coil levitated at its current height.

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.

What is a good rule of thumb for designing a simple magnetic coil? Specifically, given a circular coil of radius \(\sim 1 \mathrm{~cm},\) what is the approximate magnitude of the magnetic field, in gauss per amp per turn? (Note: \(1 \mathrm{G}=0.0001 \mathrm{~T}\).) a) \(0.0001 \mathrm{G} /(\mathrm{A}\) -turn \()\) b) \(0.01 \mathrm{G} /(\) A-turn \()\) c) \(1 \mathrm{G} /(\mathrm{A}\) -turn \()\) d) \(100 \mathrm{G} /(\mathrm{A}\) -turn \()\)

A 0.90 m-long solenoid has a radius of \(5.0 \mathrm{~mm} .\) When the wire carries a 0.20 - A current, the magnetic field in the solenoid is \(5.0 \mathrm{mT}\). How many turns of wire are there in the solenoid?

A square loop of wire with a side length of \(10.0 \mathrm{~cm}\) carries a current of 0.300 A. What is the magnetic field in the center of the square loop?
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