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Hi, my name is hafiz sharif and today we're going to solve problem 28, sorry problem 29 .8 of chapter 29 from the text book university physics with modern physics by young sears and zimanski that is the 14th edition.
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So we're just going to read the problem to begin with.
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The problem reach, there's a calculation problem first of all, and the problem says a flat circular steel loop of radius 75 centimeters is at rest in a uniform magnetic field as shown in the figure in, as shown in, sorry, as shown in an edge on view in figure e29 .8, their field is changing with time according to b of t is equal to 1 .4 tesla down the exponential of minus 0 .057 per second time and a of the problem says find an emf induced in a loop as a function of time.
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So we're going to do a first.
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I'm going to read part b and see how the problem i should get to that.
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So i'm going to write problem.
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I hope i get the hand to write.
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Just hide it a bit, going to write.
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This is problem 29.
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On it, and we have part a.
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So part a we're going to calculate the emf.
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To calculate the emf, you may want to recall the equation derived from both faraday and lenin's law.
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And this equation is simply the emf, emas for electromagnetic force, due to the changing magnetic field, is equal to the negative time derivative of the magnetic flux.
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So if the magnetic flux is not changing with time, then the emf is zero.
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But specifically in this problem, the magnetic flux is changing with time.
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So we can at least expect a no zero emf.
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So basically, when i say faraday and lennox's law, i mean, the fact that the ems is related to the time derivative of the magnetic flux comes from faradis work or faradis law, and a negative sign that precedes that derivative, it has a result of lennox's law.
03:33
So to calculate the emf, i'm going to call this equation one.
03:38
To calculate the emf, we have to take the derivative of this changing magnetic flux.
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We know that the magnetic flux itself is giving us the dot product of the magnetic field, also known as the magnetic flux density and the area vector.
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In other words, the magnetic flux is the product of the magnetic field or the magnetic flux density in the area in particular to.
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This magnetic field.
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Usually, this area has to be one of a close loop as it is in the problem.
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We know from previous physics classes that the dot product can also be defined.
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The dot product of b and a can be defined as the magnitude of b, that is the magnitude of the magnetic field, the magnitude of this area, the cosine of some angle of five, in this case, and i'm going to call this angle 5, where 5 is an angle between the area vector and the magnetic field.
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Remember that the area vector is one that points in a direction normal to the surface of the area itself.
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So to calculate this magnetic flux, we need to know three things.
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We need to know what the magnetic field is, at least the magnitude of the magnetic field.
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We need to know what the area is and we need to know what the angle fire is between the magnetic field and the area vector.
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So we begin with their magnetic field.
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The magnetic field is giving a problem as equal to 1 .4 tesla times the exponential of negative 0 .0 .0 five seven per second time so the magnetic flux is just giving the the magnetic field sorry is just giving the problem we don't have to solve for that i want to circle this as an important sorry i got the red into it i want to circle this as an important equation that we're going to need in calculating the magnetic flux the next we need the next quantity we need this the area since we the problem says that the area that their loop is made of it's made of circular steel we can calculate this area by computing the area of a cycle an area of a circle as we know when geometry is by times the square of the radius we know we know no pi.
07:38
I'm going to just use an approximate value for pi as 3 .14.
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And the radius is given a problem.
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The problem says the radius of this circular steel loop is 75 centimeters.
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So the radius is given directly.
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75 cm.
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Since the magnetic field up here is in test line.
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Is in si units.
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It will make sense for us to just converge it 75 centimeter directly to si units before going forward.
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So the converter is to as a unit we say we use something called a factor labor metal.
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The si unit for length is meter.
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So put it here because that is the target unit we want to go to and the unit we're coming from the centimeter so we're actually in the middle.
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Yeah.
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And the conversion factor is such that for every one meter we have we have 100 centimeter.
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So, centimeter can't a centimeter, and we're left with 75 times 1, that's 75, and divide 75 by 100, should be meters, and our radius is now 0 .75 meters.
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Since we now know the radius and we know that pi is a constant, we can calculate the area of this loop.
09:38
The area of the loop is pi arrow square, remember? pi for us is, sorry, i was about to write the length, i'll about to write the radius first.
09:58
Now, pi is 3 .1 .4.
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The radius is 0 .75 meters.
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You work this out, the area equals, i worked it out on my calculator previously.
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And just so that you know, i'm going to keep four decimal places, i'm going to keep four decimal places in all of my intermediate answer and round up at the end.
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It just helps me get a more precise answer.
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So if you multiply 3 .14 by 0 .75, of course, we have to square the radius.
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We're going to have an area of 1 .763 meter square.
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This is the area of our circular steel loop.
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Now we know what the area is and we know what the magnetic field is.
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We now need to figure out.
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What the angle phi is.
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To get that angle fire, angle five, angle five, angle five, we need to remember one thing.
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Now don't confuse fire with the magnetic flux.
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That's what i'm writing the substrate of b here.
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To get this angle fire, we need to look at the geometry of the problem.
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If you look at the geometry of the problem in figure e20, and even the wording of the problem, we told that from the wording of the problem that there, our steel circular loop is flat.
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So i'm trying to draw a flat line here.
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So this is our steel circular loop.
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Here is our loop.
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And from the problem you see that the magnetic field lines makes some angle with this loop.
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There is some angle here with this loop.
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From the problem.
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And this angle is, sorry, this angle is 60 degrees.
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These red lines are the magnetic fuel lines.
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Well then, the area vector, the area vector for this circular loot of area 1 .7663 points, as i told you earlier, normal to the surface of the area itself.
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So this is our area vector.
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This green vector is our area vector.
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So you see that the angle phi that we want to find is not this angle 60 degree instead.
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It is this angle.
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This is our angle five.
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Because if you look up this equation, we said this phi is the angle between nette field and the area vector.
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The magnetic field is represented by these red lines.
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The area vector is this green vector.
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So this is the angle between the green vector and the rare vector.
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This is five.
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You see that five and sixty degrees are complementary angles.
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This is to say that five plus 60 degrees was equal 90 degrees.
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Therefore, five is equal to 90 degrees.
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90 degrees minus 60 degrees.
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And our angle 5, that exists between the area vector and our magnetic field is 30 degrees.
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This is the tier parameter that we needed to go ahead and solve the problem.
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Now that we have all of these, of all of what we need, we now know what the magnetic fuel is that is given to us in the problem.
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We've calculated the area of the circular loop.
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We have figured out what the angle is between the area vector and a magnetic field vector.
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We can go ahead and calculate the magnetic flux.
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Remember this equation.
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A magnetic flux from the previous page is the product of the magnetic field, the area and cosine of angle 5.
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We soft to you all of what we just formed, we're going to have the magnitude of the magnetic field.
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It's 1 .4 tesla e.
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Bristol minus 0 .057.
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Set per second times time.
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It's going to be this.
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This whole thing is the magnetic field.
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The area the area we just formed is 1 .7663 meters square cosine of five and we just found five to be 30 degree this is the magnetic flux.
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Next we need another page if we put all of those together and do the multiplication what we get is that the magnetic flux is equal to if i did my calculation right that is my arithmetic right going to have 2 .1 .1415e is to minus 0 .1057 per second down time and of the units afterwards that unit's going to be tesla times meter square.
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We don't work on what the units would be.
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So units is our next goal.
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Our unit ends up with tesla times meter square.
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But we know that tesla itself, sorry, we know that tesla itself, tesla is weber per meter square.
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So tesla per meter square, tesla times meter square, sorry, is just weber per meter square times meter square.
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Meter square scales on middle square.
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And we're left with weber, which is indeed the si unit for magnetic flux.
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So we've indeed calculated that our magnetic flux by sub b is 2 .1 .1415.
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Weber times e.
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Reach to negative 0 .057 per second times time.
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This is the magnetic flux but remember that a of this problem did not ask us to just find the magnetic flux but a of this problem asks us to find or to calculate the emf induced in the loop as a function of time so we now go back to equation one that gives the emf compute the emf as required we know that the emf from equation 1 is equals to minus the first time derivative of the magnetic flux with respect to time.
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We have found that, we have found what the magnetic flux is.
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So we just go ahead and compute this derivative.
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It should be negative, first time derivative of what we just found.
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Objects from happens to be 2 .1415 wherever times e which 0 .0 .057 per second times time.
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Compute this derivative for the usual rules of differentiation we see that 2 .141 word is a constant.
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So it comes outside of the differentiation sign.
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And now we have to differentiate the exponential term.
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To differentiate the exponential term, the first thing you do is that you write the exponential term down by itself.
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The exponential term reproduces itself first.
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It's like you differentiate the exponential term.
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And then you differentiate the argument of the exponential term.
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The argument of the exponential term is this expression.
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So d by d of negative 0 .057 per second times time.
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There's a time term out there.
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We should not forget it.
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So this becomes minus or negative 2 .155.
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Weber.
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So we do the differentiate the derivative first.
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To differentiate minus a constant times the variable, we just put the constant out, that's going to be minus 0 .057 per second.
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And we differentiate t, we respect the t will get one.
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This happens to be the derivative term.
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This term.
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Comes here and we just write down an exponential term as it is.
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0 .057.
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Now we have to multiply minus 2 .14 .15 wherever or minus 0 .057 per second.
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To do that multiplication, negative signs go out and we're left with 0.
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Sorry we're not left with but the product of this multiplication is 0 .1 221 we will figure out the unit soon we're just going to keep the product here and write the exponential term 0 .0571 t and the units however per second.
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Now we have to remember one we ever is one fourth second.
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So weber a second is fourth second per second so we get a fourth second.
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For this we have our final result.
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We'll need a new page for that.
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So the emf becomes zero 0 .1 -221 votes f .h.
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Response to the initial term which is era to negative 0 .0 .0.
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0 .057 a second dance t.
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Or we can say the emf is 1 -22 to 122 .1 miller volts times exponential term, which is 0 .057 per second time.
28:52
So we go back to the statement of the problem where acts by impact a to find the emf induced in the loop as a function of time.
29:01
Indeed, we have found the emf induced in a loop as a function of time.
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To be very specific, we can write this e as a function of t.
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This is the answer to part a of this problem.
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This is an answer to the part a of this problem.
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I hope everything is clear.
29:36
You should be able to see that.
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Indeed, what we have calculated is an emf.
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Because emf is actually a voltage or a driving voltage.
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And the unit of voltage is four.
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And you can see here that our final answer has you know what...