Magnetics 4 Question 4
5. The dipole moment of a circular loop carrying a current $I$ is $m$ and the magnetic field at the centre of the loop is $B _1$. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is $B _2$. The ratio $\frac{B _1}{B _2}$ is
(2018 Main)
(a) $\frac{1}{\sqrt{2}}$
(b) 2
(c) $\sqrt{3}$
(d) $\sqrt{2}$
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Answer:
Correct Answer: 5. (d)
Solution:
- $m=I \times \pi R^{2}$
$$ \begin{aligned} & 2 m=I \times \pi\left(R^{\prime}\right)^{2} \Rightarrow R^{\prime}=\sqrt{2} R \\ & B=\frac{\mu _0 I}{2 \pi R} \Rightarrow B \propto \frac{1}{R} \Rightarrow \frac{B _1}{B _2}=\frac{R^{\prime}}{R}=\sqrt{2} \end{aligned} $$
