Magnetics Ques 4
- A square loop is carrying a steady current $I$ and the magnitude of its magnetic dipole moment is $m$. If this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop (in A-m) will be
(2019 Main, 10 April II)
(a) $\frac{4 m}{\pi}$
(b) $\frac{3 m}{\pi}$
(c) $\frac{2 m}{\pi}$
(d) $\frac{m}{\pi}$
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Answer:
Correct Answer: 4.( a )
Solution:
- Key Idea Magnetic dipole moment of a current carrying loop is
$ m=I A\left(A-m^2\right) $
where, $I=$ current in loop and $A=$ area of loop. Let the given square loop has side $a$, then its magnetic dipole moment will be
$ m=I a^2 $
Then, wire length will be same in both areas,
$ \Rightarrow \quad 4 a=2 \pi r \Rightarrow r=\frac{4 a}{2 \pi}=\frac{2 a}{\pi} $
Hence, area of circular loop formed is, $A^{\prime}=\pi r^2$
$ =\pi\left(\frac{2 a}{\pi}\right)^2=\frac{4 a^2}{\pi} $
Magnitude of magnetic dipole moment of circular loop will be
$ m^{\prime}=L A^{\prime}=I \frac{4 a^2}{\pi} $
Ratio of magnetic dipole moments of both shapes is,
$ \frac{m^{\prime}}{m}=\frac{I \cdot \frac{4 a^2}{\pi}}{I a^2}=\frac{4}{\pi} \Rightarrow m^{\prime}=\frac{4 m}{\pi}(\mathrm{A} \cdot \mathrm{m}) $
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