1. We see the edge of a 55 turn flat circular coil of magnetic dipole moment 0.1
ID: 1869960 • Letter: 1
Question
1. We see the edge of a 55 turn flat circular coil of magnetic dipole moment 0.194 A-m2 in magnitude. The eye (not ours) looks along the coil's axis, "seeing" a counterclockwise current of 8.8 A. Find the coil's area vector. (The "DIRECTION', here is relative to the paper, not relative to the eye, so answer "right ()" or "left ()") VECTOR FQUATION USED SOLUTION ANSWER MAGNITUDE col DIRECTION es an angle of 60.0 with a flat surface. Thus it makes an angle of 90.0-60.0 A uniform magnetic fie 30.0° with the normal to the surface. The area of the surface is 6.66 × 10-6 m2. The resulting magnetic flux through the surface is 4.44 nWb. Calculate the magnitude of the magnetic field to three significant figures. EQUATION USED CONE EQUAL SIGN) SOLUTION ANSWER . A straight wire segment is 1.27 cm long and carries a substantial current of 99 A into the paper (C ) through a huge uniform external magnetic field of 45 T out of the paper (O). Find the force exerted by the field on the segment. VECTOR EQUATION USEDD SOLUTION ANSWER I find the magnitude using--()()(-)(sin-) I use (Put symbols in the five blanks above.) (Write a number above for .) MAGNITUDE DIRECTIONExplanation / Answer
1.
Given
number of turns is n= 55 , magnetic dipole moment is me = 0.194 A.m2
current i = 8.8 A
we know that the magnetic moment is mue = n*I*A = 55*8.8*A
A = mue/(n*I)
A = 0.194/(55*8.8) m^2
A = 0.000400 m^2
Area vector is A = 0.000400 m^2
the dirction of the magnetic moment with counterclock wise current in the coilis to the right
2.
Area is A = 6.66*10^-6 m2
magnetic flux phi = 4.44 nWb
the angle made by the field with Area (normal) = 30 degrees
we know that the magnetic flux is ph i = B*A cos theta
B = phi/(A cos theta)
B = (4.44*10^-9)/(6.66*10^-6 cos30) T
B = 0.0007698003589 T
B = 769.800*10^-6 T
3.
length of the wire is L = 1.27 cm = 0.0127 m
current i = 99 A directed into the page
magnetic field is B = 45 T ( out of the page)
the force on the wire is F = i *L*B sin theta
F = 0.0127*99*45 sin 90 N
F = 56.5785 N
the force direction is upward
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