What is the magnitude of the torque exerted on the coil by a uniform magnetic field B = 0.630 T directed in the positive x direction when the current is I = 1.20 A in the direction shown?

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A rectangular coil consists of N = 130 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 θ = 30.0° with the x axis (see figure below).

image

(a) What is the magnitude of the torque exerted on the coil by a uniform magnetic field B = 0.630 T directed in the positive x direction when the current is I = 1.20 A in the direction shown?
N · m

(b) If you are looking downward from the positive y direction, what is the expected direction of rotation of the coil?

    
asked Feb 19, 2012 in Physics by Joey33 ~Expert~ (1,216 points)
    

1 Answer

+1 vote
 
Best answer

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 = 130 

a = 0.400 m and b = 0.300 m

Area = a*b =  0.4m * 0.3m = 0.12 m^2

B = 0.630 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 = (130) * (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 = ((130) * (1.2 A) * (0.12 m^2)) * ( 0.630 T) * sin(60)

Torque = 10.21 N*m

answered Feb 24, 2012 by pokemonmaster ~Expert~ (3,856 points)
selected Feb 24, 2012 by Joey33



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