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Part EWhat is the magnetic field strength at point c? Express your answer to two slgnificant figures and Include the approprlate unlts:B =ValueUnitsSubmnitBequesLAn...

Question

Part EWhat is the magnetic field strength at point c? Express your answer to two slgnificant figures and Include the approprlate unlts:B =ValueUnitsSubmnitBequesLAntwet

Part E What is the magnetic field strength at point c? Express your answer to two slgnificant figures and Include the approprlate unlts: B = Value Units Submnit BequesLAntwet



Answers

What are the magnetic field strength and direction at points a to c in FIGURE EX33.14? (FIGURE CAN'T COPY)

We have be at a equals B for the top on the bottom. Added together. So that will be you. Not I. Over two pi d times consigned 45. I had minor signed 45 jihad for the top and for the bottom is consigned 45. I had plus signed 45 jihad. That's a sign. So putting all the numbers in this will cancel out the jihad will cancel out. It will be two times tend to them. Negative four. I had ESA at point B magnetic field just given by mu, not I over two pi d. I had times two on that is four times 10 to the 94. I had ESA at point C. It is so that is just too be Mandy Phil, that point c will be such that this would be a positive sign and this will be a negative sign. So this will equal toe

Hello students in this question we have to find the magnetic field at point C. Okay so we have the two disciples that is over and 02. Okay. And the magnetic movement of the first one is and when equals two am not and for the second one M two it is equal to 16. I'm not. So this open is situated at the actual point. So that magnetic field of even will be equals to the you know two divided by 45. And this I'm going to ever buy. These are not Q. Okay so everyone is equals to this. I am not some you not divided by four path blurb. I am not aware they are not cute. And this Magnetic field will be in the direction of the Apple movement. So this will be towards the right. Okay now talking about the magnetic field B2 which is due to the photo. So this will this is on the equatorial position. Sorry. Here we will have another factor too. Yeah. To do excel situation. Okay this is exhale, this is equatorial. So um you know to develop before by and this is um to develop our mhm dash is cute. Okay so you not divided by 45 and M. Two which is equal to 16 and not ever be our dish which is to our not so too are not to the power three. And this disciple after this magnetic field after solving is equals two mu not divided by 45. Yeah who am not where they are not cute and this will be opposed to the double movement because it is in the equatorial situation. So the net magnetic field at once, she will be equals two B one plus B two. Both are equal in magnitude. Okay? But oppose it in the direction. So the net result will be equals to zero. Okay, so this is the answer for the problem. Okay, so from the given options option C. Is the correct answer here. Thank you.

Right In the given problem here, there is the magnetic field within a circular region. Both radius is given as our that with a lot. This is the radius of this magnetic field which is directed into the plane of paper, and it is given having given as having a magnitude be, then the three observation points at which we have to put the positive charge Q and then have to find electrostatic force acting on it. The first point is above the center three of this such region circular region. The magnetic well, this is the point understands our from the center then towards right. This is another point B that is also at a distance R and the third point is see itself at which we have to find the electrostatic forces acting on the charged particles. So, first of all, before finding the electrostatic forces because we know expression for electrostatic forces, the product of Q into even Easter electric field. So, first of all, we should find electric fields at these three points A B and C Corp. Which we will use the relation for the your math induced. As for Faraday's law of electromagnetic induction, that is given as the negative of the time rate of change of magnetic flux linked through a region and then that e. M f is actually potential difference And potential difference has given us the negative of line integral of electric field or one an imperial blue. So yeah, it will become cancelling this negative sign. This is differentiation of being to aid because the magnetic flux is given as the product of magnetic field with the area is equal to the product of electric field. At these points A and B multiple advice circumstance which is to buy our so taking this a means area as a constant out This is D B by DT. You have this magnetic field is increasing and the rate of D B bye deedee so in to deliver DT is equal to e into two by oh for area this is pi r squared. But here the magnetic field the magnetic flux which we will consider should be only within this circular region. So for area, we will consider the radius to be small are only so get diseased the area of the small region by our square into db by DT is equal to e into two by our or finally we can say electric field he's given us are by two into dp by TT So the force will be given. Ask you into e or we can say this is Q are by two into db by eating so this liver the expression for the force experienced by the charged particle. Now we have to find the directions of these forces also so to find it, first of all, s magnetic field is increasing inward in your direction. So as per lenses law, the indie used magnetic field should oppose this magnetic field. So the induced magnetic field should be increasing outward means coming out of the plane of paper and, as you know, magnetically comes out of the North Pole and we know North Wall is formed by counter clockwise current or counterclockwise electric field. Mhm. So we conclude that the direction of electrically induced at these three points is counter clockwise means the direction of electrical at a will be towards left at B, it will be towards upward in upward direction and that will be the single with the direction of force, so direction of force experienced by positive charge at 0.8 will be towards left at point B. It will be upward in the plane of paper, but there will be no electric field. Hence no electrostatic force at point C. When the child particle is put at one C, there will be no electric field and hence no electrostatic force. Thank you.

High in the given problem. There are a loop, There is a loop in the form of the combination. Well, two arbs like this. These arcs are making an angle off 1 20 degree at the center. The smaller arc is having a radius, smaller is equal to 20 Centimeter and uh bigger arc is having a radius. Capital art is equal to 30 centimetre. The current carried by these two arcs, the same current will be carried by them. Whether these are the arcs or the street segments, so the same current will be passing through them, whose magnitude is given as 12.0 mp. Here. Now we have to find the net magnetic field here at the center of these two ox for which we will use a concept that as we know magnetic field at the center of a circular coil, it's given as B is equal to new, not I by tour. And that circle. If we consider this to be the circle. The total angle made by the circumference at the center will be two point. So the magnetic field due to the complete angle means to buy angle is do not I by two are. So. If you consider an arc, a smaller arc making an angle theater at the centre, then using unitary methods, the expression for magnetic field at the center. Do you do? Arc will be given by P. Is equal to theater by to buy or we can say this is 3 60 degree into the normal magnetic will means mu not I buy to our. And then this expression will be used in order to find the magnetic field due to the two are shaped portions. But in order to find the magnetic field due to these straight lines, we will use the concept. We will use Barton several slot which says a magnetic will do to the small current carrying element ideal at a point B whose position veterans making an angle theta. That magnetic will smaller magnetic is given by You're not by four by into ideal sine theta by our square. So if that point he comes here exactly along the length of this conductor angle, theta will come out to be zero. Means in that case the magnetic field will be zero. Hence we conclude the magnetic field at the center. See due to these two straight segments will be zero. So magnetic field magnetic fields at the center due to stripped Segments will be zero. So the net magnetic will is only because of these two are shaped portions. Now, when more concept here we will use as in this smaller arc, the current is in a clockwise direction. So this phase of the smaller coil will behave like south pole is the direction of magnetic field is into the plane of paper. And due to this bigger arc the magnetic field will be outward and as we know, the magnetosphere will be stronger due to this shorter portion, the smaller rock. So the magnetic field which is into the plane of paper that will be having larger magnitude. Hence the net magnetic field will be given by Tita by 360 into mu. Not I by two are smaller minus theta by 3 60 degree into um you know, try again, Divided by two capital are For theatre, this is 120° Divided by 3 60°.. Um you note by two and I all the things can be taken As a common out, leaving behind one by AR -1 by capital are in the Bracket. No plugging in all other known values. First of all, this is one by three into form, you know, this is four pipes Into 10 for -7, divided by two into for current, this is 12 MP er one by Our means 20 cm means 28 to 10, 8 for minus two m minus one by 30 centimetre, or 30 into 10 for -2 m. So here it is two times here it is four times. So finally this is Eight pi into 10 days for -7 and taking the standard for -2. Also in the denominator is comin out, LCM for 20 and 30 will be 60, so this is 3 -2 and that is Tesla. So finally this magnet, if it is eight in to tend bar -5 into one by 60 what we can see, This is 1.3 into 10 H 4 -6 Tesla. So finally we can say this is 1.3 Micro Tesla, and the direction is into the plane of paper, which becomes the answer for the given problem. Thank you.


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