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Ch 27: Magnetic Field and Magnetic Forces

Chapter 27, Problem 27

An open plastic soda bottle with an opening diameter of 2.5 cm is placed on a table. A uniform 1.75-T magnetic field directed upward and oriented 25° from vertical encompasses the bottle. What is the total magnetic flux through the plastic of the soda bottle?

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Hey everyone. So this problem is working with magnetic flux, let's see what they give us and what they're asking for. So we know that there is a uniform magnetic field of a certain strength, it is directed at an angle they give us from the Z directions. That's important. Um We have a water picture that's made out of glass and it has a circular plastic top. So we have two different materials here and then they give us the top radius. We know the picture with the glass or the picture with the plastic top lies within the magnetic field and it's on top of a horizontal table. So we need to determine the total magnetic flux that's passing through the picture glass. So the first thing we need to do here is recognize that we have two different materials. So the total flux is going to be the flux of the glass plus the flux of the plastic. We can also recall that the total flux and a closed surface is zero. We know it's closed. We have a top on top of the picture and so the total The magnitude of the total flux is going to be zero. So we can rewrite this as the flux of the glass is equal to minus the flux of the picture. Let's recall that the equation for flux is given as B. A. Times Kassian data. Now they're asking us for the flux of the glass, but we really have no way of calculating the area of the glass picture through this relationship though. We can calculate the area of the plastic top and then find the flux of the plastic top, which then gives us the flux of the glass picture. So let's calculate the flux through the plastic top. We have the magnetic field, the area of the plastic top and the cosine of theta. So take this term by term. The magnetic field strength is given as 1.5 1.15 Tesla. The area of the top, what's called? The area of the circle is just pi R squared. So I'm going to rewrite our in terms of meters that 0.7 m squared, that comes out to 1.54 times to the -2 m squared. And then uh See our data was given to us in the equation as 37°. So from here we can just plug all of that into our equation. And we come out with 1.4, 1 times 10 to the -2 Levers, going back up to this original equation. They're asking us for the flux through the glass. We know that's the opposite of the flux through the plastic top. And so the flux through the glass picture is minus 1.41 times 10 to the negative second Weber's. We go and look up our potential answers and that aligns with answer a that's all we have for this problem. We'll see you in the next video
Related Practice
Textbook Question
A flat, square surface with side length 3.40 cm is in the xy-plane at z = 0. Calculate the magnitude of the flux through this surface produced by a magnetic field B=(0.200 T)i+(0.300 T)j-(0.500 T)k .
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Textbook Question
A circular area with a radius of 6.50 cm lies in the xy-plane. What is the magnitude of the magnetic flux through this circle due to a uniform magnetic field B = 0.230 T (b) at an angle of 53.1° from the +z-direction?
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Textbook Question
A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an angle of 30.0° above the horizontal. What must the magnitude of the magnetic field be to produce a flux of 3.10x10^-4 Wb through the surface?
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Textbook Question
A 150-g ball containing 4.00x10^8 excess electrons is dropped into a 125-m vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude 0.250 T and direction from east to west. If air resistance is negligibly small, find the magnitude and direction of the force that this magnetic field exerts on the ball just as it enters the field.
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Textbook Question
Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 155 V/m and the magnetic field is 0.0315 T. The ions next enter a uniform magnetic field of magnitude 0.0175 T that is oriented perpendicular to their velocity. (a) How fast are the ions moving when they emerge from the velocity selector? (b) If the radius of the path of the ions in the second magnetic field is 17.5 cm, what is their mass?
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Textbook Question

A 150-V battery is connected across two parallel metal plates of area 28.5 cm^2 and separation 8.20 mm. A beam of alpha particles (charge +2e, mass 6.64x10^-27 kg) is accelerated from rest through a potential difference of 1.75 kV and enters the region between the plates perpendicular to the electric field, as shown in Fig. E27.29 <IMAGE>. What magnitude and direction of magnetic field are needed so that the alpha particles emerge undeflected from between the plates?

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