B = 0.7i T
Magnetic force on a current carrying conductor
Fm = I (l X B)
Magnitude of the forces
F1 :
l= -aCos(31) i - aSin(31) j = -0.086 i - 0.052 j
F1 = 19*11( (-0.086 i - 0.052 j) X 0.7 i ) T = 7.6k N
F2 : : l = -0.2 k
F2 = 19*11 ( -0.2 k X 0.7 i) = -29.26 k N
l3 = +aCos(31) i + aSin(31) j
F3= 19*11( (0.086 i + 0.052 j) X 0.7 i ) T = -7.6k
l4 = 0.2 k
F4 = 19*11 ( 0.2 k X 0.7 i) = 29.26 k N
b) magnetic moment of a current loop = o IA
Magnetic moment of the coil
| M | = NI A = 19*11* (0.2*0.1) = 4.18 A-sq.m - magnitude.
direction is parallel to A and CCW to the current
M = 4.18 ( -aSin(31) i + aCos(31) j )
= -0.22 i + 0.36 j A-m2
Torque on the coil
= M X B = (-0.22 i + 0.36 j ) X 0.7 i
= -0.252 k N-m
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