Answer
given,
current (I) = 16 mA
circumference of the circular loop (S)= 1.90 m
Magnetic field (B)= 0.790 T
S = 2 π r
1.9 = 2 π r
r = 0.3024 m
a) magnetic moment of loop
M= I A
M=
M=
M=
b) torque exerted in the loop



Explanation:
The 11Ω, 22Ω, and 33Ω resistors are in parallel. That combination is in series with the 4Ω and 10Ω resistors.
The net resistance is:
R = 4Ω + 10Ω + 1/(1/11Ω + 1/22Ω + 1/33Ω)
R = 20Ω
Using Ohm's law, we can find the current going through the 4Ω and 10Ω resistors:
V = IR
120 V = I (20Ω)
I = 6 A
So the voltage drops are:
V = (4Ω) (6A) = 24 V
V = (10Ω) (6A) = 60 V
That means the voltage drop across the 11Ω, 22Ω, and 33Ω resistors is:
V = 120 V − 24 V − 60 V
V = 36 V
So the currents are:
I = 36 V / 11 Ω = 3.27 A
I = 36 V / 22 Ω = 1.64 A
I = 36 V / 33 Ω = 1.09 A
If we wanted to, we could also show this using Kirchhoff's laws.
Answer:
can't tell if this is question, it is not written correctly
Explanation:
Electrical conductivity is the measure of a material's ability to allow the transport of an electric charge. Its SI is the siemens per meter, (A2s3m−3kg−1) (named after Werner von Siemens) or, more simply, Sm−1. It is the ratio of the current density to the electric field strength.