Solveeit Logo

Question

Question: Explain with the help of a labeled diagram the distribution of the magnetic field due to a current t...

Explain with the help of a labeled diagram the distribution of the magnetic field due to a current through a circular loop. Why is it that if a current-carrying coil has NN turns, the field produced at any point is n times as large as that produced by a single turn?

Explanation

Solution

To answer this question, we have to first draw a diagram for a current through a loop. To answer the second part, we should remember the formulae of the magnetic field Biot Savart’s law for the magnetic field produced by a circular loop. We can write the magnetic field for one loop and then the magnetic field for NN turns and compare.

Formulae used:
B=μ0NI4πRB = \dfrac{{{\mu _0}NI}}{{4\pi R}}
Here BB is the magnetic field produced due to a circular loop, NN is the number of turns, II is the current carried by the loop, μ0{\mu _0} is the permeability of free space and RR is the radius of the loop.

Complete answer:

In the above figure, BB is the magnetic field that has magnetic field lines produced due to a circular loop and II is the current carried by the loop and its direction in the loop. we can clearly see that using the right-hand thumb rule we can easily predict the direction of magnetic field lines.
In figure (a)(a) , we are trying to find the direction using the right-hand rule, where the thumb points toward the direction of current flow and the curling of fingers show the direction of the magnetic field.
In figure (b)(b) , shows the direction of the magnetic field.
We know that
B=μ0NI4πR\Rightarrow B = \dfrac{{{\mu _0}NI}}{{4\pi R}}
Here BB is the magnetic field produced due to a circular loop, NN is the number of turns, II is the current carried by the loop, μ0{\mu _0} is the permeability of free space and RR is the radius of the loop.
For one turn the amount current flowing in the loop will be II
So the magnetic field will be
B1=μ0I4πR\Rightarrow {B_1} = \dfrac{{{\mu _0}I}}{{4\pi R}}
But for NN turns, the current will also increase NN times. So the magnetic field will be,
BN=μ0NI4πR=NB1\Rightarrow {B_N} = \dfrac{{{\mu _0}NI}}{{4\pi R}} = N{B_1}
So we know that the magnetic field of a current-carrying loop increases with an increase in the number of turns of the loop.

Note: While calculating the magnetic field for the object, make sure to use the correct formulae. For example, here we need to use the formula for the magnetic field for a loop. So no need to use the formulae of the magnetic field for a wire. Also while finding the direction of the magnetic field for any current-carrying wire or loop, make sure to use the right-hand rule correctly.