I believe the answer is b) slowly heating the surface
Answer:
The track's angular velocity is W2 = 4.15 in rpm
Explanation:
Momentum angular can be find
I = m*r^2
P = I*W
So to use the conservation
P1 + P2 = 0
I1*W1 + I2*W2 = 0
Solve to w2 to find the angular velocity
0.240kg*0.30m^2*0.79m/s=-1kg*0.30m^2*W2
W2 = 0.435 rad/s
W2 = 4.15 rpm
To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:
![\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}](https://tex.z-dn.net/?f=%5Coint%20%5Cvec%7BB%7D%5Cvec%7Bdl%7D%20%3D%20%5Cmu_0%20%5Cepsilon_0%20%5Cfrac%7Bd%5CPhi_E%7D%7Bdt%7D)
Where,
B= Magnetic Field
l = length
= Vacuum permeability
= Vacuum permittivity
Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to
![B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}](https://tex.z-dn.net/?f=B%282%5Cpi%20r%29%20%3D%20%5Cmu_0%20%5Cepsilon_0%20%5Cfrac%7Bd%28EA%29%7D%7Bdt%7D)
Recall that the speed of light is equivalent to
![c^2 = \frac{1}{\mu_0 \epsilon_0}](https://tex.z-dn.net/?f=c%5E2%20%3D%20%5Cfrac%7B1%7D%7B%5Cmu_0%20%5Cepsilon_0%7D)
Then replacing,
![B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}](https://tex.z-dn.net/?f=B%282%5Cpi%20r%29%20%3D%20%5Cfrac%7B1%7D%7BC%5E2%7D%20%28%5Cpi%20r%5E2%29%20%5Cfrac%7Bd%28E%29%7D%7Bdt%7D)
![B = \frac{r}{2C^2} \frac{dE}{dt}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7Br%7D%7B2C%5E2%7D%20%5Cfrac%7BdE%7D%7Bdt%7D)
Our values are given as
![dE = 2150N/C](https://tex.z-dn.net/?f=dE%20%3D%202150N%2FC)
![dt = 5s](https://tex.z-dn.net/?f=dt%20%3D%205s)
![C = 3*10^8m/s](https://tex.z-dn.net/?f=C%20%3D%203%2A10%5E8m%2Fs)
![D = 0.440m \rightarrow r = 0.220m](https://tex.z-dn.net/?f=D%20%3D%200.440m%20%5Crightarrow%20r%20%3D%200.220m)
Replacing we have,
![B = \frac{r}{2C^2} \frac{dE}{dt}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7Br%7D%7B2C%5E2%7D%20%5Cfrac%7BdE%7D%7Bdt%7D)
![B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B0.220%7D%7B2%283%2A10%5E8%29%5E2%7D%20%5Cfrac%7B2150%7D%7B5%7D)
![B =5.25*10^{-16}T](https://tex.z-dn.net/?f=B%20%3D5.25%2A10%5E%7B-16%7DT)
Therefore the magnetic field around this circular area is ![B =5.25*10^{-16}T](https://tex.z-dn.net/?f=B%20%3D5.25%2A10%5E%7B-16%7DT)
The force exerted by the magnetic in terms of the magnetic field is,
![F\propto B](https://tex.z-dn.net/?f=F%5Cpropto%20B)
Where B is the magnetic fied strength and F is the force.
Thus, if the magnetic A has twice magnetic field strength than the magnet B,
Then,
![B_A=2B_B](https://tex.z-dn.net/?f=B_A%3D2B_B)
Thus, the force exerted by the magnet B is,
![\begin{gathered} F_B\propto B_B \\ F_B\propto\frac{B_A}{2} \\ F_B=\frac{F_A}{2} \\ F_B=\frac{100}{2} \\ F_B=50\text{ N} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20F_B%5Cpropto%20B_B%20%5C%5C%20F_B%5Cpropto%5Cfrac%7BB_A%7D%7B2%7D%20%5C%5C%20F_B%3D%5Cfrac%7BF_A%7D%7B2%7D%20%5C%5C%20F_B%3D%5Cfrac%7B100%7D%7B2%7D%20%5C%5C%20F_B%3D50%5Ctext%7B%20N%7D%20%5Cend%7Bgathered%7D)
Thus, the force exerted by the magnet B on magnet A is 50 N.
The force exerted by the magnet A exerts on the magnet B is exactly 100 N as given.
Hence, the option B is the correct answer.