North compass to batteries A bar magnet with magnetic dipole moment o.5 A m2 lle

Question

North compass to batteries A bar magnet with magnetic dipole moment o.5 A m2 lles on the negative x-axis, as shown in the diagam. A compass is located at the orlgin. Magnetic North is in the negatez direction. Between the bar magnet and the compass is a coll of wire of radius .5 cm, connected to batteries not show n n the picture. The dstance from the center of the coll to the center of the compass s 9.4 cm. The distance from the center of the bar magnet to the center of the compass is 21.5 cm. A steady current of 1 037 amperes runs through the coil. Conventional current runs counter-clockwise in the coil when viewed from the location of the compass Despite the prasence of the coil and the bar magnet, the compass points North. which end of the bar magnet is closer to the compass?- Select Chock all of the following statements that are corract: It isn't necessary to take the magnetic field of the magnet into account, since the coil is in between the magnet and the compass The magnetic dipole moment of the coil is equal to the magnetic dipole moment of the compass. Because not all parts of the rnagnel dre the sar ne distance from he uonpass treatinu Lhe magnet dS d dipole ucated al the center of the magnet is dn dpprox mdLion. At the location af tha comp s, tha ner magnetic field is equal to the magnetic field of the Farth. | At the location of the compass, the magnetic field due to the magnet is equal in magnitude to the magnetic field due to the coil. How many turns of wire are in the coll? number of turns-

Explanation / Answer

B magnetwill be opposite of Bcoil

B magnet=(o/4)x(2/r3)

Bcoil=NxBloop

N=Bmagnet/Bloop

Given =0.5 Am2

r=21.5cm=0.215m

Bmagnet=((4x10-7)/4)x(2/r3)

          =(10-7x2x0.5)/(0.215)3

          =1.006x10-5

Bcoil=(o2R2I)/(4(Z2+R2)3/2)

o=4x10-7

Bcoil=(10-7x2xR2xI)/( (Z2+R2)3/2)

Given R=0.015m

I=1.037amp

Z=0.094m

Bcoil=(10-7x2x3.14x(0.015)2x1.037)/[(0.094)2+(0.015)2]3/2

=(1.4652x10-10)/(8.6251x10-4)

=1.698x10-7

Number of turns N=Bmagnet/Bcoil

N=(1.006x10-5)/(1.698x10-7)

N=59.246

The above all statements are correct

Therefore number of turns N=59.246

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