MHT CET · Physics · Magnetic Effects of Current
A steel wire of length ' \(\ell\) ' has a magnetic moment 'M'. It is then bent into a semicircular arc. The new magnetic moment is
- A \(2 \mathrm{M} / \pi\)
- B \(\mathrm{M}\)
- C \(\mathrm{M} \times \ell\)
- D \(\mathrm{M} / \ell\)
Answer & Solution
Correct Answer
(A) \(2 \mathrm{M} / \pi\)
Step-by-step Solution
Detailed explanation
(C)
When wire is bent in the form of semicircular arc then, \(\mathrm{l}=\pi \mathrm{r}\) \(\therefore\) The radius of semicircular arc, \(\mathrm{r}=1 / \pi\)
Distance between two end points of semicircular wire \(=2 \mathrm{r}=\frac{21}{\pi}\) \(\therefore\) Magnetic moment of semicircular wire \(=m \times 2 r=m \times \frac{2 l}{\pi}=\frac{2}{\pi} m l\)
But \(\mathrm{ml}\) is the magnetic moment of straight wire i.e., \(\mathrm{ml}=\mathrm{M}\)
\(\therefore\) New magnetic moment \(=\frac{2}{\pi} \mathrm{M}\)

When wire is bent in the form of semicircular arc then, \(\mathrm{l}=\pi \mathrm{r}\) \(\therefore\) The radius of semicircular arc, \(\mathrm{r}=1 / \pi\)
Distance between two end points of semicircular wire \(=2 \mathrm{r}=\frac{21}{\pi}\) \(\therefore\) Magnetic moment of semicircular wire \(=m \times 2 r=m \times \frac{2 l}{\pi}=\frac{2}{\pi} m l\)
But \(\mathrm{ml}\) is the magnetic moment of straight wire i.e., \(\mathrm{ml}=\mathrm{M}\)
\(\therefore\) New magnetic moment \(=\frac{2}{\pi} \mathrm{M}\)

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