A rectangular coil of wire, 22.0 cm by 35.0 cm and carrying a current of 1.50 A,

Question

A rectangular coil of wire, 22.0 cm by 35.0 cm and carrying a current of 1.50 A, is oriented with the plane of its loop perpendicular to a uniform 2.50-T magnetic field, as shown in ihe figure. (a) Calculate the net force and torque that the magnetic field exerts on the coil. (b) The coil is rotated through a 60.0degree angle about the axis shown, with the left side coming out of the plane of the figure and the right side going into the plane. Calculate the net force and torque that the magnetic field now exerts on the coil. (c) What is the change in potential energy of the coil when it is rotated from 0degree to 60.0degree?

Explanation / Answer

a) Fnet = 0 , Net Torque that magneticfield exerts on the coil is 0

b) If the coil is rotated by 600 about the axis then Fnet is 0

Torque = iABsin600 = 1.5*(0.22*0.35)*2.50*sin600

= 0.250064 N-m

c) Change in Potential Energy of the coil when it is rotated from 00 to 600 is Uf - Ui

   = (-p.B)f - (-p.B)i

= iABcos600 + iABcos00

= 0.433125 J

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