A 51.0-turn circular coil of radius 5.40 cm can be oriented in any direction in

Question

A 51.0-turn circular coil of radius 5.40 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.530 T. If the coil carries a current of 25.9 mA, find the magnitude of the maximum possible torque exerted on the coil. A rectangular coil consists of N = 150 closely wrapped turns and has dimensions a = 0.400 m and b = 0.300 m. The coil is hinged along the y axis, and its plane makes an angle theta = 30.0degree with the x axis (figure). (a) What is the magnitude of the torque exerted on the coil by a uniform magnetic field B = 0.450 T directed in the positive x direction when the current is I = 1.20 A in the direction shown? If you are looking downward from the positive y direction, what is the expected direction of rotation of the coil?

Explanation / Answer

mmmAA) torque=rxF=NrFsin(theta)

for max torque sin(theta)=1

torque=rF=NBIA

GIVEN :N=51,r=5.4cm,B=0.530T,I=25.9mA, Area=pir^2=3.14*5.4*5.4*10^-4

torque=51*(0.530)*(25.9*10^-3)*(3.14*5.4*5.4*10^-4)=0.0064N

B)

Torque = magnetic moment X magnetic field (*where X = cross product)

Torque = magnetic moment * magnetic field * sin(phi)

where,

magnetic moment of coil = N*I*A (*where N =number of closely wrapped turns, I=current, A=area)

we are given the following values in the question:

N = 150

a = 0.400 m and b = 0.300 m

Area = a*b = 0.4m * 0.3m = 0.12 m^2

B = 0.450 T

I = 1.20 A

theta = 30 degrees

phi = 60 degrees (* it is the angle between the magnetic moment and the magnetic field 90-30 = 60 )

so the magnetic moment = N*I*A

magnetic moment = (150) * (1.2 A) * (0.12 m^2)

Furthermore plugging this back into the formula for torque we get:

Torque = magnetic moment * magnetic field * sin(theta)

Torque = ((150) * (1.2 A) * (0.12 m^2)) * ( 0.450 T) * sin(60)

Torque = 8.41 N*m

Anti clock wise

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