Answer:
Magnetic flux through the loop is 1.03 T m²
Explanation:
Given:
Magnetic field, B = 4.35 T
Radius of the circular loop, r = 0.280 m
Angle between circular loop and magnetic field, θ = 15.1⁰
Magnetic flux is determine by the relation:
....(1)
Here A represents area of the circular loop.
Area of circular loop, A = πr²
Hence, the equation (1) becomes:
![\Phi=B\pi r^{2} \cos \theta](https://tex.z-dn.net/?f=%5CPhi%3DB%5Cpi%20r%5E%7B2%7D%20%5Ccos%20%5Ctheta)
Substitute the suitable values in the above equation.
![\Phi=4.35\times\pi (0.28)^{2} \cos 15.1](https://tex.z-dn.net/?f=%5CPhi%3D4.35%5Ctimes%5Cpi%20%280.28%29%5E%7B2%7D%20%5Ccos%2015.1)
= 1.03 T m²
A proton travels through a constant magnetic field in the negative y-direction while moving in the negative x-direction. The proton will be subject to a magnetic pull that is directed into the page. Option B is correct.
<h3>What is the right-hand thumb rule?</h3>
Hold a current-carrying conductor in your right hand with your thumb pointing in the direction of the current then wrap your fingers around the conductor and orient them in the direction of the magnetic field lines.
A proton travels through a constant magnetic field in the negative y-direction while moving in the negative x-direction.
The proton will be subject to a magnetic pull that is directed into the page.
Hence, option B is correct.
To learn more about the right-hand thumb rule refer to the link;
brainly.com/question/11521829
#SPJ1
Answer:
Explanation:
Given
Length of rope ![L=55\ m](https://tex.z-dn.net/?f=L%3D55%5C%20m)
Weight of rope ![W=4040\ N](https://tex.z-dn.net/?f=W%3D4040%5C%20N)
weight density
Work done to lift rope 33 m
![W=\int_{0}^{33}\lambda hdh](https://tex.z-dn.net/?f=W%3D%5Cint_%7B0%7D%5E%7B33%7D%5Clambda%20hdh)
![W=\int_{0}^{33}73.45hdh](https://tex.z-dn.net/?f=W%3D%5Cint_%7B0%7D%5E%7B33%7D73.45hdh)
![W=73.45\left [ \left ( \frac{h^2}{2}\right )\right ]^{33}_0](https://tex.z-dn.net/?f=W%3D73.45%5Cleft%20%5B%20%5Cleft%20%28%20%5Cfrac%7Bh%5E2%7D%7B2%7D%5Cright%20%29%5Cright%20%5D%5E%7B33%7D_0)
A. <span>I .................
</span>