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At the equator, the earth’s field is essentially horizontal;near the north pole, it is nearly vertical. In between, the anglevaries. As you move farther nort...

Question

At the equator, the earth’s field is essentially horizontal;near the north pole, it is nearly vertical. In between, the anglevaries. As you move farther north, the dip angle, the angle of theearth’s field below horizontal, steadily increases. Green turtlesseem to use this dip angle to determine their latitude. Suppose youare a researcher wanting to test this idea. You have gathered greenturtle hatchlings from a beach where the magnetic field strength is50 μT and the dip angle is 56∘. You

At the equator, the earth’s field is essentially horizontal; near the north pole, it is nearly vertical. In between, the angle varies. As you move farther north, the dip angle, the angle of the earth’s field below horizontal, steadily increases. Green turtles seem to use this dip angle to determine their latitude. Suppose you are a researcher wanting to test this idea. You have gathered green turtle hatchlings from a beach where the magnetic field strength is 50 μT and the dip angle is 56∘. You then put the turtles in a 1.1 m diameter circular tank and monitor the direction in which they swim as you vary the magnetic field in the tank. You change the field by passing a current through a 150-turn horizontal coil wrapped around the tank. This creates a field that adds to that of the earth. What current should you pass through the coil, to produce a net field in the center of the tank that has a dip angle of 62∘ ? Express your answer with the appropriate units.



Answers

At the equator, the earth's field is essentially horizontal; near the north pole, it is nearly vertical. In between, the angle varies. As you move farther north, the dip angle, the angle of the earth's field below horizontal, steadily increases. Green turtles seem to use this dip angle to determine their latitude. Suppose you are a researcher wanting to test this idea. You have gathered green turtle hatchlings from a beach where the magnetic field strength is $50 \mu \mathrm{T}$ and the dip angle is $56^{\circ} .$ You then put the turtles in a 1.2-m-diameter circular tank and monitor the direction in which they swim as you vary the magnetic field in the tank. You change the field by passing a current through a 100 -turn horizontal coil wrapped around the tank. This creates a field that adds to that of the earth. What current should you pass through the coil, and in what direction, to produce a net field in the center of the tank that has a dip angle of $62^{\circ} ?$

Hi that even problem. The magnetic field at the center off the oil is equal toe Earth's magnetic field. So toe find the value off Earth's magnetic field. We just find the value off magnetic field at the center of this current carrying coil, which is given by the expression you, not an arm by you are here number off France in the coil. Mhm 200. The radios off this coil, which is half of the diameter and diameter hair, is given as 1 m. So this becomes 1 m by 2015 speaker and the current in this guy is 0.215 MPs, Yes, the magnitude of this magnetic comes out to be for new note. This is poor pile into 10 dish for minus seven Tesla meter for India multiplied by 200 then multiplied by current, which is 0.215 and here divided by two times off radius which is to 0.5 m. The units which will be cancer help uh, and here and meter so leaving behind Just Tesla. Yes, a magnet Europe. This magnetic really comes out of the 1.8 into 10 days for minus four. That's not or we can say this is be visible. Toe, 108 micro Tesla, which is the answer for the given problem. So at this Prince Arts magnetic field should have value off 108 Micro Tesla. Thank you.

Hi. In the given problem, oil one is in a vertical plane. Yeah, this is called one Carrying current in such a manner that it's magnetic field is a long no direction. Then the call to is in a horizontal plane carrying current in such a direction so that its magnetic field will be vertically upward like this. So here, this is B one. And here this is B two now arts magnetic field has been given as this be at an angle off 61 degree from north in a poor direction. Both the coils are having equal number off turns, which is 43. The radio soft botha coils is also same, which is 1.0 m arts. Magnetic field has been given us 5.6 in 10 days to part minus five s lock. Now we have to find the current passing through the coil one and passing through coil to as there is no not magnetic field at the center. Uh, the coy's. So the component off arts magnetic field along the X axis means along horizontal will be equal in magnitude. Toby even is a magnetic field. You toe the coil one and opposite in direction. Similarly, the component off the vertical component off Earth's magnetic field will be having the same magnitude as that off the magnetic field you to call to and opposite in direction. So if the current's in oil one and oil too are supposed to be, I won and I do respectively. Then this be one should be equal toe. Horizontal component off Earth's magnetic field means B E cause 61 degree. Hence, we can say for Beaven, this is new not, and one I won, divided by two are one is ableto for B. This is 5.6 to 10 dish bar minus five cause 61 degree. Hence the value off if you put the values off all known terms here for new notices 45 in tow tended bar minus seven. Number of trans 43 currently have toe find radius. This is 1.0 m is equal to 5.6 into 10. Dish par minus five cause 61 degrees. Hence the value off. This current I one we become two, multiplied by 5.6 in tow. 10 dish par minus five cause 61 degree divided by four body in tow. 10 dish par minus seven in 2 43 which comes out Toby equal to 1.0 into 10. Which part of minus seven. And, yeah, which is first answer for this given problem. Now we have to find the current passing through call to also for which the magnetic field off call to be, too will be equal toe. Vertical component off Earth's magnetic field means B E sign 61 degree. So it says new not and do I to divided by you are too visible. Toe 5.6 into 10 days. Four minus five into sign 61 degree. So here also the value I, too, will become 2 to 5.6 in tow. 10 days far minus five Sign. 61 degree divided by 45 in tow. 10 days par minus seven into 43. So the value of this current I toe comes out. Toby. 1.8 in tow. 10 for minus seven m b r. And this is the second answer for a given problem. Thank you

Any problem that start with me. Writing the radius of the earth is probably going to be a good one. So question number seventy six wants us to imagine the very rial scenario of wrapping a single wire around the earth. Um, where we want to try to wreck, create the Earth's magnetic field of roughly one G. So this is what we're trying to get to one times ten to the minus four Tess Lem near the poles. And we're recreation. That by a single current loop, As I said and we want to know what current that Luke is going to need to carry S o. I think again, this is a fair, feasible question. We used the standard equation for, ah, loop of wire and rearrange it for the current wishes. What we're most interested in, so, too, are he times the field strength to find it. By this probability. Ah, and then we plug everything in into our calculation, and we see that it is going to require a puny one times ten to the nine amps of current. So hopefully we have a superconducting wire or just a really thick one, because that is a huge amount

Hi. The given problem there is a circular coil kept vertically in such a manner that the normal drawn at explain is pointing do you east? And there will be as this is a current carrying call. So there will be a magnetic field that would be be service Group C. The magnetic field will be along the axis of the soil and as this disease. So here there will be not in which there will be the horizontal component of earth's magnetic field. Due to which when we put a compass here at the center of this school, the compass until will not show along the access. But This will be the direction of compass at an angle of 60° north of east. Now for this coil, the number of terms in the circular coil is an is equal to hunt it. It's three years. He's given us our is equal to 20 cm. Or we can say this is 0.20 meter. The current passing through this coil is 0.400 mps. Direction of campus is 60 degree north of east now to find the horizontal component of earth's magnetic field in the first part of the problem. First of all, we find magnetic will be easy at the center of the current carrying coil using the expression you not and I buy two are So plugging in unknown values form you know this is four pi into 10 days par -7 islam meter for Ambien for the number of crunch, this is 100 and for current disease. 0.400 and beer. And here this ampere will be cancelled here we can cancel the Zambia And divided by two. Are there R. is the radius which is 0.20 meters. So this meter will also be cancelled. Hence finally, this magnetic field at the center of this circular current can enjoy comes out to be well 0.56 Into tended bar -5 Tesla. Now, in order to find this beach, we use the expression for this angle dan B to. So here this 10 better he's given us For 10 Dickler upon base be edge upon b. c. As the Bh is unknown. So expression for this bh horizontal component of what's magnetically we become Bc times 10. Beat up instances 12.56 Into Tended Bar -5. Yes, this was -5, multiplied by 10 60 degree. So finally this horizontal component of what's magnetic field Comes out to be 2.18 into tended bar minus for Tesla. Answer for the first part of this problem. Now, in the second part of the problem, if you draw this drive around here, suppose this is the north direction in which this is the horizontal component of Earth's magnetic field. Its vertical component is vertically downward. BB. So the total magnetic field Earth's magnetic field is supposed this and this angle here is known as angle off inclination delta. Know this angle of inclination has been given in the problem as 50 five degree. So using the expression for horizontal component V. H. Is equal to this. Be the total magnetic field of Earth B. cause 55°.. So the earth's total magnetic field will be given by B. H. Divided by cause 55 degrees Means this is 2.18 into 10 days par -4, divided by The value of course 55° 0.574. So finally this Earth's magnetic field here comes out to be 3.8 into tentage par minus for Tesla approx. And this is the answer for that 2nd part of this problem. Thank you.


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