The magnetic moment is -
M = 5.5 x 
A
.
We have current carrying single circular loop placed in a magnetic field
parallel to the plane of the loop.
We have to determine the Magnetic moment of the loop.
<h3>What is the formula to calculate the magnetic moment of loop?</h3>
The formula to calculate the magnetic moment of the loop is -
M = NIA
where -
N - Number of turns
I - Current in the loop
A - Area of the loop
According to the question, we have -
I = 17mA = 0.017A
Circumference (C) = 2 meters
B = 0.8 T
Now -
C = 2
2
r = 2
r = 1
r = 
r = 0.32
Using the formula -
M = NIA
M = NI
M = 3.14 x 1 x 0.017 x 0.32 x 0.32
M = 3.14 x 4 x 0.017
M = 5.5 x A
Hence, the magnetic moment is -
M = 5.5 x A
To solve more questions on Magnetic moments, visit the link below-
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Here’s a good photo to reference when converting in the metric system.
Each time you move down a step you move the decimal to the right, each time you move up a step you move the decimal to the left.
We are going from 1.2 kg or kilograms, which is at the very top left of the ladder. To get to mg or milligrams, we would have to make six jumps, so we’d move the decimal over six times.
1.2 > 12. > 120. > 1200. > 12000. > 120000. > 1200000.
So our final answer would be 1,200,000mg.
Answer:
Explanation:
From the law of conservation of energy
Energy lost by the spring, W=Kinetic energy gained, KE+Potential energy gained, PE+Work done by friction, Fr
The required distance from A to B is
Answer:
(A) Th = 818.6 K
(B) Qh = 14211.7 J
Explanation:
efficiency (n) = 0.537
temperature of cold reservoir (Tc) = 379 K
heat rejected (Qc) = 6580 J
(A) find the temperature of the hot reservoir (Th)
n = 1 - 
0.537 = 1 - 
= 1 - 0.537 = 0.463
Th = 
Th = 818.6 K
(B) what amount of heat is put into the engine (Qh) ?
from 
Qh = 6580 ÷ 
Qh = 14211.7 J
