GOAL Calculate a magnetic torque on a loop of current PROBLEM A circular wire lo
ID: 2030386 • Letter: G
Question
GOAL Calculate a magnetic torque on a loop of current PROBLEM A circular wire loop of radius 1.00 m is placed in a magnetic field of magnitude 0.500 T. The normal to the plane of the loop makes an angle of 30.0° with the magnetic field (Fig. a). The current in the loop is 2.00 A in the direction shown. (a) Find the magnetic moment of the loop and the magnitude of the torque at this instant. 2.00 m .00 m r- 100 (b) The same current is carried by the rectangular 2.00-m by 3.00-m coil with three loops shown in figure b. Find the magnetic moment of the coil and the magnitude of the torque acting on the coil at that instant. 2.00 m 3,00 m STRATEGY For each part, we just have to calculate (a) A circular current loop in an extemal magnetic field the area, use it in the calculation of the magnetic moment, and multiply the result by B sin 0. Altogether, this process amounts to substituting values into the correct equation. (b) A rectangular current loop in the same field. (c) (used in exercise). SOLUTION (A) Find the magnetic moment of the circular loop and the magnetic torque exerted on it. First, calculate the enclosed area of the circular loop (1.00 m)2 3.14 m2 Calculate the magnetic moment of the loop IAN-(2.00 A)(3.14 m2)(1) 6.28 A.m2 - sin -(6.28 A-m2)(0.500 T)(sin 30.0°) Now substitute values for the magnetic moment, magnetic field, and into the equation to right 1.57 N m (B) Find the magnetic moment of the rectangular coil and the magnetic torque exerted on it: Calculate the area of the coil: A-L × H = (2.00 m)(3.00 m) 6.00 m2 Calculate the magnetic moment of the coil: IAN-(2.00 A)(6.00 m2)(3) 36.0 A m2 Substitute values into the equation - sin -(0.500 T)(36.0 A-m2)(sin 30.0°) 9.00 N mExplanation / Answer
We know that
u = i A = i (pi r^2)
u = 2.1 x 3.14 x 1.23^2 = 9.98 A-m^2
Hence, u = 9.98 A-m^2
tau = u B sin(theta)
tau = 9.98 x 0.52 x sin30 = 2.59 N-m
Hence, tau = 2.59 N-m
b)u = n i A
A = 2 x 3 = 6 m^2
u = 3 x 2.1 x 6 = 37.8 A-m^2
Hence, u = 37.8 A-m^2
tau = u B sin(theta)
tau = 37.8 x 0.52 x sin30 = 9.83 N-m
Hence, tau = 9.83 N-m
c)u = i A
A = 1/2 b h = 0.5 x 1.55 x 3.15 = 2.44 m^2
u = 2.1 x 2.44 = 5.124
n = 2 ; u = 2 x 5.124 = 10.25 A-m^2
Hence, u = 10.25 A-m^2
tau = u B sin(theta)
tau = 10.25 x 0.52 x sin30 = 2.67 N-m
Hence, tau = 2.67 N-m
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