Question

(1 point) A rectangular coil has 19 turns and sides of length a = 10 cm and c = 20 cm. The coil is pivoted about the z axis and its plane is at 0 = 31' to a magnetic field B = 0.7iT, as shown in the figure above. If I = 11 A, find: a) the force on each side. F =( k)N F2 =1 k) F: =( k) F4 = k)N b) What is the magnetic moment of the coil? M =( i+ k) Am c) What is the torque on the coil? T=( k)N.m

Answer 1

B = 0.7i T

Magnetic force on a current carrying conductor

F_m = 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

F_{2 :} : 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 = $\mu$ _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

$\vec{\tau }$ = M X B = (-0.22 i + 0.36 j ) X 0.7 i

= -0.252 k N-m