A rectangular current carrying 54 turn coil, as shown in the figure, is pivoted

Question

A rectangular current carrying 54 turn coil, as shown in the figure, is pivoted about the z axis. The current in the coil is I = 1.70 A. (a) If the wires in the z = 0 plane make an angle theta = 36 degree with the y axis, what angle does the magnetic moment of the coil make with the unit vector i^? (b) Write an expression for n^in terms of the unit vectors i^and j^, where n^is a unit vector in the direction of the magnetic moment. (c) What is the magnetic moment of the coil? (d) Find the torque on the coil when there is a uniform magnetic field B vector = 1.3 T j^in the region occupied by the coil. (e) Find the potential energy of the coil in this field. (The potential energy is zero when theta = 0.)

Explanation / Answer

part a)

theta=36º

part b )

n = nxi + nyj

n = cos36 i - sin36 j

n = 0.809 i - 0.587 j

part c )

mu = NIA*n

mu = 54 * 1.70 * ( 8 *10^-2 x 5 *10^-2) ( 0.809 i - 0.587 j )

mu = (0.2970 m^2) i - (0.2155 m^2) j

part d )

t = mu x B

t =  [(0.2970 m^2) i - (0.2155 m^2) j ]x (1.3T)j

t = (0.3861 N-m ) k

part e )

U = -mu . B

U = -[ (0.2970 m^2) i - (0.2155 m^2) j ] . (1.3 T)j

U = 0.280 J

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