Question

A 2 m 2m 2 m

The magnetic field varies according to the tesla equation B = (ti- (3x ^ 2) * (t) * (- j) -x * t (k) in the given area. Resistance density of Ac and AD is neglected. Resistance density of AB and BC is 1ohm / m, BD and DC are 2ohm / m What is the current passing through each branch.

Answer 1

Given, magnetic field is
B = i(t) + j(3x^2*t) - k (xt)
Now, from the figure,
we can see, the x component of field passes through the triangle, y component through loop in zx plane and z component through xy plane

Also, given, AC and AD dont have any resistance
AB has resistance density of 1 ohm / m
BC has resistance density of 1 ohm / m
BD and DC has resistance density 2 ohm / m

hence, EMF generated in the triangular loop is V1 = -d(t*0.5*2*2)/dt = -2 V
-ve sign is clockwise EMF
i1 = V1/R1
R1 = 4 ohm
i1 = -2/4 = -0.5 A

Similiarly, for y component of magnetic field
V2 = -d(phi2)/dt
d(phi2) = 3x^2t*2*dx = 6x^2*t*dx
phi2 = integral(d(phi2)) = 2t*8 = 16t
hence
V2 = -d(16t)/dt = -16 V (clockwise)
R2 = 8 ohm
i2 = -16/8 = -2 A

For z component of magnetic field
V3 = -d(phi3)/dt
d(phi3) = -xt*2*dx = -2xtdx
phi3 = -4t
V3 = 4 V (counterclockwise)
R3 = 4 + 4 + 4 + 4 = 16 Ohm
i3 = 4/16 = 0.25 A

from superposition
current in AC = i1 = 0.5 A from A to C
current in AD = 2 - 0.5 = 1.5 A from A to D
current in DC = 0.25 + 0.5 = 0.75 A from C to D

current in AB = 2 A clockwise
current in BD = 2 + 0.25 = 2.25 A from D to B
current in BC = 0.25 A counterclockwise