KCET · Physics · Magnetic Effects of Current
A circular coil carrying a certain current produces a magnetic field \(B_{0}\) at its centre. The coil is now rewound so as to have 3 turns and the same current is passed through it. The new magnetic field at the centre is
- A \(\frac{B_{0}}{9}\)
- B \(9 B_{0}\)
- C \(\frac{B_{0}}{3}\)
- D \(3 B_{0}\)
Answer & Solution
Correct Answer
(B) \(9 B_{0}\)
Step-by-step Solution
Detailed explanation
The magnetic field produced at the centre of the circular coil carrying current is given by
\(B V=\frac{\mu_{0} N I}{2 r}\)
For one turn
\(\begin{aligned}&N=1 \\&B_{0}=\frac{\mu_{0} I}{2 r}\end{aligned}\)
As the coil is rewound
\(r^{\prime} =\frac{r}{3}, N^{\prime}=3\)
\(\therefore B^{\prime} =\frac{\mu_{0} I \times 3}{2 \times\left(\frac{r}{3}\right)}=\frac{9 \mu_{0} I}{2 r}=9 B_{0}\)
\(B V=\frac{\mu_{0} N I}{2 r}\)
For one turn
\(\begin{aligned}&N=1 \\&B_{0}=\frac{\mu_{0} I}{2 r}\end{aligned}\)
As the coil is rewound
\(r^{\prime} =\frac{r}{3}, N^{\prime}=3\)
\(\therefore B^{\prime} =\frac{\mu_{0} I \times 3}{2 \times\left(\frac{r}{3}\right)}=\frac{9 \mu_{0} I}{2 r}=9 B_{0}\)
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