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1 ptsQuestion 14couinterthe Ranths You vant magnetic fveld Iurns with Tadius ofrsop T)usingMasnrtic ficld ofacoil ( 2mm} with 50Q0vouneed che cciltocrcale masnctic...

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1 ptsQuestion 14couinterthe Ranths You vant magnetic fveld Iurns with Tadius ofrsop T)usingMasnrtic ficld ofacoil ( 2mm} with 50Q0vouneed che cciltocrcale masnctic fiekl Calculatcihe amperage thz coil Use _ dccima pl:ices in yauranswver:aan mapnitudethe middle of

1 pts Question 14 couinterthe Ranths You vant magnetic fveld Iurns with Tadius ofr sop T)using Masnrtic ficld ofacoil ( 2mm} with 50Q0 vouneed che cciltocrcale masnctic fiekl Calculatcihe amperage thz coil Use _ dccima pl:ices in yauranswver: aan mapnitude the middle of



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Conceptual Questions A solenoid with ends marked $A$ and $B$ is suspended by a thread so that the core can rotate in the horizontal plane. A current is maintained in the coil so that the electrons move clockwise when viewed from end $A$ toward end $B .$ How will the coil align itself in Earth's magnetic field?

In this problem on the topic of induction were given two coils, the first of which has an conductance of 25 million Henry's and 100 tons while is coiled too has an induction of 40 million Henry's and 200 tons. The coils are fixed in place and have a mutual induction of three million hendry's a current in the coil one is six million peers, changing at a rate of four mps per second. And we want to find the magnetic flux fire one to the self induced PMF appearing in coil one, the magnetic flux 5 to 1. And the mutually induced E. M. F. Appearing in coil too. Now, firstly the flux appearing in coil one is the induct in L one times the current I one divided by the number of turns in coil one, which is N. One. And so this is 25 millie Henry's times the current of six million peers, divided by the number of turns 100, which gives us the Flux to be one 0.5 micro weapons. Yeah, the magnitude of the self induced E. M. F. In coil one is L one times direct. Which the current is changing. In coil one D I one D. T. So this is 25 million Henry's times the rate of change of current given to before mps per second. And so we get this MF the south India CMF and coil one to be one Times 10 to the power two million balls. Now in coil to we find the flux 5-1 to be the mutual induction. M Times the current through. I won over the number of turns in Quail two and 2. Yeah, This is a mutual induction of three millie Henry's times a Current income in coil one of 6 million peers Divided by the number of turns in Coil two, which is 200. This gives a flux of 90 Nano Weber's. Yeah. And for part D, the mutually induced e m f. Absalon to one is equal to the mutual inductions, m Times The I one DT. And so this is three molly Henry's times for mm pia's per second, Which gives an m. f. of 12 milli volts.

In this problem, while in the form of circle wired in the form of circle, Where in the form of cell kill of one turn of one and with magnetic field with magnetic field at center it will be at center it B when the same when the same it's bent in each banned in lieu of Anton's, then the new, then the new magnetic field, then the new magnetic field. It is any square times of auditing so I can write. The value of BDS is equal to any square B. So option B. Each correct answer.

So here for part A, we can find the the flux and coil one. So essentially we can say the magnetic flux and Coyle. One would be equal to else of one ice of one divided by and someone This is equaling 25,000,000 Henry's multiplied by 6.0 milli amps, divided by 100. And so this would give us 1.5 Micro Weber's Ah, for part B, we confined the magnitude of the self induced E M F. So essentially we can say the IMF because they AMs of one equals the inducted, its ally else of one, multiplied by the change in the current someone with respect to time. So this would be again 25. Miller Henry's multiplied by 4.0 and peers per second, and this is giving us 1.0 times 10 to the second Millet bolts. Now, for part C, we can say in coil to we're doing the exact same thing so we can say that of magnetics. Yield in the magnetic fields up to would be equal to M m times i sub one divided by and so to this is equaling again. 3.0 Miller Henry's multiplied by 6.0 milli amps, divided by 200 instead of 100. And this is equaling 90. Nana whoever. So this would be your answer for part c. Your answer for a part b Your answer for a part. Ay and finally, four part D. We want to find the induced IMF in essentially coil, too. This would be equal to M times the change in the current in coil one with respect the time solicit again be 3.0 Miller Henry's multiplied by 4.0 and peers per second. This is giving us 12 Mila volts. This would be our answer for a party That is the end of the solution. Thank you for watching.

As we know that I am musical to N I A, Which is equal to 200 multiplication, one multiplication by Jari Square, Which is required. 200 multiplication by multiplication than multiplication. Then to the power -2 police query. Strategical to buy mm meters squared. So according to the option option eight, option H, correct answer. But this problem, option eight, correct answer for this problem. I hope you understand the solution. I am repeating the concept once again. 1st. I just write the formula I am is equal to N I A. After that, I just put all the value and just solve it to get the final and storage option E.


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