9. At a location near the equator, the earth\'s magnetic field is horizontal and
ID: 2029796 • Letter: 9
Question
9. At a location near the equator, the earth's magnetic field is horizontal and points north moving vertically upward from the ground. What is the direction of the magnetic force (a), East I and points north. An electron is South (c) North (d) The magnetic force is zero (e) West that acts on the electron? 10. A square current-carrying loop is placed in a uniform magnetic field B withR the plane of the loop parallel to the magnetic field (see the drawing). The dashed line is the axis of rotation. The magnetic field exerts (a) (b) a net force and a net torque on the loop. neither a net force nor a net torque on the loop. a net force, but not a net torque, on the loop. a net torque, but not a net force, on the loop. Axas thro ugh center of loop 11. Two circular coils of current-carryi 0.097 m, Ni-150 turns, and carries a current of Ii-2.5 A. T of l-7.5 A. What is the radius of the second coil? (Hint: Magnetic moment of a coil- NIA, area Aar ng wire have the same magnetic moment. The first coil has a radius of The long straight wire carries a current of 5.0 A along the z direction. e) Calculate the magnitude of the magnetic field at a distance of 0.050 m om the wire. ( B-ul 2m, ,-4 0-7 N / A2 )Explanation / Answer
11.
Magnetic moment of a coil is given by:
M = N*I*A
If both coils have equal magnetic moment, then
M1 = M2
N1*I1*A1 = N2*I2*A2
A = pi*r^2
N1*I1*pi*r1^2 = N2*I2*pi*r2^2
N1*I1*r1^2 = N2*I2*r2^2
r2 = r1*sqrt ((N1/N2)*(I1/I2))
Using given values:
r2 = 0.097*sqrt ((150/220)*(2.5/7.5))
r2 = 0.046 m
radius of second coil = 0.046 m
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