Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The magnitude is
, the direction is into the page
Explanation:
From the question we are told that
The current is 
The radius of arc bc is 
The radius of arc da is 
The length of segment cd and ab is = 
The objective of the solution is to obtain the magnetic field
Generally magnetic due to the current flowing in the arc is mathematically represented as
Here I is the current
is the permeability of free space with a value of 
r is the distance
Considering Arc da

Where
is the angle the arc da makes with the center from the diagram its value is 
Now substituting values into formula for magnetic field for da
![B_{da} = \frac{4\p *10^{-7} * 12}{4 \pi (0.20)}[\frac{2 \pi}{3} ]](https://tex.z-dn.net/?f=B_%7Bda%7D%20%3D%20%5Cfrac%7B4%5Cp%20%2A10%5E%7B-7%7D%20%2A%2012%7D%7B4%20%5Cpi%20%280.20%29%7D%5B%5Cfrac%7B2%20%5Cpi%7D%7B3%7D%20%5D)
![= \frac{10^{-7} * 12}{0.20} * [\frac{2 \pi}{3} ]](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B10%5E%7B-7%7D%20%2A%2012%7D%7B0.20%7D%20%2A%20%5B%5Cfrac%7B2%20%5Cpi%7D%7B3%7D%20%5D)

Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to da is into the pages of the paper
Considering Arc bc

Substituting value
![B_{bc} = \frac{4 \pi *10^{-7} * 12}{4 \pi (0.30)} [\frac{2 \pi}{3} ]](https://tex.z-dn.net/?f=B_%7Bbc%7D%20%3D%20%5Cfrac%7B4%20%5Cpi%20%2A10%5E%7B-7%7D%20%2A%2012%7D%7B4%20%5Cpi%20%280.30%29%7D%20%5B%5Cfrac%7B2%20%5Cpi%7D%7B3%7D%20%5D)

Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to bc is out of the pages of the paper
Since the line joining P to segment bc and da makes angle = 0°
Then the net magnetic field would be



Since
then the direction of the net charge would be into the page
<span>Since youc oncetrate all your force directly towards the moment arm it means that you push it at an angle of your force is directed to the left or the right and I bet that it must be 90</span> degrees to the bar. Obviuosly, if you are about to push it you will do it straight up but not in a zig zag way. In other words, it should be perpendicular to the arm because the<span> torque can be produced only if force is applied at a constant index (90).
Hope that helps! Regards.</span>
Possibly, if you have list of densities and you have to match it. I can't think of any other scenarios in which it would be able to.
Hope I helped! :)
So we want to know what is the purpose of a lanyard attached to a safety switch. So in case the operator falls overboard a safety switch is installed and connected to the operators hand or waist. Which ever is more practical. This safety switch turns off the motor.