Equatorial currents are primarily westward. This is because the dominant current in the northern hemisphere has a clockwise direction, while the southern hemisphere has a counterclockwise direction. When these two currents meet at the equator, a common westward current exists.
Answer:
The height is 
A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )
Explanation:
From the question we are told that
The height is 
The angle of the slope is 
According to the law of conservation of energy
The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

Where I is the moment of inertia which is mathematically represented as this for a sphere

The angular velocity
is mathematically represented as

So the equation for conservation of energy becomes
![mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2](https://tex.z-dn.net/?f=mgh_s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5B%5Cfrac%7B2%7D%7B5%7D%20mr%5E2%20%5D%5B%5Cfrac%7Bv%7D%7Br%7D%20%5D%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]](https://tex.z-dn.net/?f=mgh_s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%5B%5Cfrac%7B2%7D%7B5%7D%20%2B1%20%5D)
![mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]](https://tex.z-dn.net/?f=mgh_s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%5B%5Cfrac%7B7%7D%7B5%7D%20%5D)
![gh_s =[\frac{7}{10} ] v^2](https://tex.z-dn.net/?f=gh_s%20%3D%5B%5Cfrac%7B7%7D%7B10%7D%20%5D%20v%5E2)

Considering a circular hoop
The moment of inertial is different for circle and it is mathematically represented as

Substituting this into the conservation equation above
![mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2](https://tex.z-dn.net/?f=mgh_c%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%28mr%5E2%29%5B%5Cfrac%7Bv%7D%7Br%7D%20%5D%20%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
Where
is the height where the circular hoop would be released to equal the speed of the sphere at the bottom



Recall that 


Substituting values

Good afternoon.
We have:

The function of velocity:

For
t = 5 s:
Answer:
(a) 0.345 T
(b) 0.389 T
Solution:
As per the question:
Hall emf, 
Magnetic Field, B = 0.10 T
Hall emf, 
Now,
Drift velocity, 

Now, the expression for the electric field is given by:
(1)
And

Thus eqn (1) becomes
where
d = distance
(2)
(a) When 

(b) When 
