Most directly on equador and least directly at poluses
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
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For rotational equilibrium of the door we can say that torque due to weight of the door must be counter balanced by the torque of external force

here weight will act at mid point of door so its distance is half of the total distance where force is applied
here we know that

now we will have


so our applied force is 72.5 N
It's actually Friction.
I just did the test and got it right.