12 V is the f.e.m.
![\epsilon](https://tex.z-dn.net/?f=%5Cepsilon)
of the battery. The potential difference that is applied to the motor is actually the fem minus the voltage drop on the internal resistance r:
![\epsilon - Ir](https://tex.z-dn.net/?f=%5Cepsilon%20-%20Ir)
this is equal to the voltage drop on the resistance of the motor R:
![RI](https://tex.z-dn.net/?f=RI)
so we can write:
![\epsilon - Ir = RI](https://tex.z-dn.net/?f=%5Cepsilon%20-%20Ir%20%3D%20RI)
and using
![r=0.0305~\Omega](https://tex.z-dn.net/?f=r%3D0.0305~%5COmega)
and
![R=0.055~\Omega](https://tex.z-dn.net/?f=R%3D0.055~%5COmega)
we can find the current I:
D is the answer. Bc the pointy
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=![16 \times 10^{-3} \times \pi \times r^2](https://tex.z-dn.net/?f=16%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%20%5Cpi%20%5Ctimes%20r%5E2)
M=![16 \times 10^{-3} \times \pi \times 0.3024^2](https://tex.z-dn.net/?f=16%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%20%5Cpi%20%5Ctimes%200.3024%5E2)
M=![4.59 \times 10^{-3}\ A m^2](https://tex.z-dn.net/?f=4.59%20%5Ctimes%2010%5E%7B-3%7D%5C%20A%20m%5E2)
b) torque exerted in the loop
![\tau = M\ B](https://tex.z-dn.net/?f=%5Ctau%20%3D%20M%5C%20B)
![\tau = 4.59 \times 10^{-3}\times 0.79](https://tex.z-dn.net/?f=%5Ctau%20%3D%204.59%20%5Ctimes%2010%5E%7B-3%7D%5Ctimes%200.79)
![\tau = 3.63 \times 10^{-3} N.m](https://tex.z-dn.net/?f=%5Ctau%20%3D%203.63%20%5Ctimes%2010%5E%7B-3%7D%20N.m)
Geometrically, a screw can be viewed as a narrow inclined plane wrapped around a cylinder. Like the other simple machines a screw can amplify force; a small rotational force (torque) on the shaft can exert a large axial force on a load.
The answer would be D. <span>The tortoise moved at a constant velocity throughout the race; the hare stopped to rest periodically.
Hope this helped. Good luck!</span>