Solveeit Logo

Question

Question: The ratio of magnetic field and magnetic moment at the centre of the current carrying circular loop ...

The ratio of magnetic field and magnetic moment at the centre of the current carrying circular loop is xx. When both the current and radius is doubled the ratio will be
A.x8A.\dfrac{x}{8}
B.x4B.\dfrac{x}{4}
C.x2C.\dfrac{x}{2}
D.2xD.2x

Explanation

Solution

We will use the relationship of magnetic field at the centre of a current carrying loop and magnetic moment at the centre of the current carrying circular loop. The ratio of magnetic field and magnetic moment will be calculated in order to solve the problem. Magnetic field is defined as the space around the magnet within which the magnetic force is applicable and magnetic moment represents its strength and orientation.
Formula Used:
We are going to use the following relations to get the correct answer:-
B=μoi2aB=\dfrac{{{\mu }_{o}}i}{2a} and M=iπa2M=i\pi {{a}^{2}}

Complete answer:
We will know about the magnetic field and magnetic moment first.
Magnetic field is defined as the space around the magnet within which the magnetic force is applicable and magnetic moment is defined as the strength and orientation of a magnet or a current carrying conductor.
We suppose a current carrying conductor with current ii having magnetic field BB and magnetic moment MM, aa is the distance from the centre.
Then magnetic field is given by the following formula:-
B=μoi2aB=\dfrac{{{\mu }_{o}}i}{2a} ………………. (i)(i) where μo{{\mu }_{o}} is magnetic permeability.
And magnetic moment is given by the following formula:-
M=iπa2M=i\pi {{a}^{2}} ………………… (ii)(ii)
Now ratio of magnetic field and magnetic moment is given as follows:-
x=BM=μoi2aiπa2x=\dfrac{B}{M}=\dfrac{\dfrac{{{\mu }_{o}}i}{2a}}{i\pi {{a}^{2}}}
x=μo2πa3\Rightarrow x=\dfrac{{{\mu }_{o}}}{2\pi {{a}^{3}}} …………….. (iii)(iii)
Now it is given that current and radius both get doubled. That is ii becomes 2i2i and radius aa becomes 2a2a. and therefore, equation (iii)(iii) becomes as follows:-
x1=μo2π(2a)3{{x}_{1}}=\dfrac{{{\mu }_{o}}}{2\pi {{(2a)}^{3}}} …………….. (iv)(iv)
x1=μo2π(8a3)\Rightarrow {{x}_{1}}=\dfrac{{{\mu }_{o}}}{2\pi (8{{a}^{3}})}
From the equation (iii)(iii) and putting it in (iv)(iv) we get,
x1=x8{{x}_{1}}=\dfrac{x}{8}

Hence, option (A)(A) is correct.

Note:
We should take care of the fact that both magnetic field and magnetic moment are different parameters with different formulas. Magnetic field is the region of magnetism and magnetic moment represents the strength and orientation of the magnetic field. Correct use of the formulae should be done.