Obtain the expression for magnetic field along the axis of toroid

B = μ_o(NI)/(2πr)

Explanation:

A torroid ideally should consist of a long conducting wire that is wound tightly around a torus which is made of a non-conducting material.

Now, when a steady current is passed through the toroid, the magnetic induction (B) in it's interior of the will be tangent to any circle that's concentric with it's axis and also has the same value on this particular circle.

If we further assume that this torroid has N turns of wire and I as the current in its coil, it means that like the Amperian loop, we will choose a circle of radius (r) that is concentric with the axis of the torroid. This means that the line integral of the magnetic field surrounding the American loop will be;

∮B.dl = ∮Bdl = B∮dl

The Integral operation gives;

B∮dl = B(2πr) - - - (1)

Furthermore, the total current that's enclosed by the Amperian loop is given by;

I_enclosed = current through each turn × number of turns enclosed by the loop

= I × N

I_enclosed = NI - - - (eq 2)

From amperes law in a vacuum, we have;

B∮dl = μ_o(I_enclosed)

Plugging in the respective equivalent expressions earlier outlined, we have;

B(2πr) = μ_o(NI)

B = μ_o(NI)/(2πr)

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.