Question

How does doubling the number of turns in a toroidal coil affect the value of magnetic flux density?
(A) Four times
(B) Eight times
(C) Half
(D) Double

Answer 1

Hint: The magnetic flux density of a toroidal coil is directly related to the number of turns of the toroid. Investigate what happens to magnetic flux by doubling the number of turns in the relation.

Formula used : In this solution we will be using the following formulae;
 $ B = \dfrac{{\mu NI}}{{2\pi r}} $ where $ B $ is the magnetic flux density (or magnetic field) of a toroidal coil or torus, $ I $ is the current flowing in the coil, $ N $ is the number of turns of the torus, and $ r $ is the radius of the torus.

Complete step by step answer:
We asked what happens to the magnetic flux density of a toroidal coil, if the number of coils of it is doubled. To do so, we shall recall the magnetic flux density of the toroid within the toroid. This is given as
 $ B = \dfrac{{\mu NI}}{{2\pi r}} $ where $ B $ is the magnetic flux density (or magnetic field) of a toroidal coil or torus, $ I $ is the current flowing in the coil, $ N $ is the number of turns of the torus, and $ r $ is the radius of the torus.
Hence, we can write that
 $ B = kN $ where $ k $ is a constant.
Now, if we doubled the number of turns, we have that
 $ {B_2} = k\left( {2N} \right) $ , which is
 $ {B_2} = 2kN $
 $ \Rightarrow {B_2} = 2B $
Hence, the magnetic flux density doubles also.
The correct option is, hence, D.

Note:
For clarity, although not specifically mentioned, this is the magnetic flux density at the centre of the turns of the toroid, i.e. which runs around the circumference of the torus, through the area encompassed by the turns of the toroid. The magnetic field outside these turns is always equal to zero no matter the number of turns or current in the toroid.