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A long insulated copper wire is closely wound as a spiral of $N$ turns. The spiral has inner radius $a$ and outer radius $b$. The spiral lies in the XY-plane and a steady current $I$ flows through the wire. The Z-component of the magnetic field at the centre of the spiral is
Options:
- A
- B
- C
- D
Solution:
1071 Upvotes
Verified Answer
The correct answer is:
If we take a small strip of $d r$ at distancer from centre, then number of turns in this strip would be
$$
d N=\frac{N}{b-a} d r
$$
Magnetic field due to this element at the centre of the coil will be
$$
\begin{aligned}
& d B=\frac{\mu_0(d N) I}{2 r}=\frac{\mu_0 N I}{2(b-a)} \frac{d r}{r} \\
\therefore \quad & B=\int_{r=a}^{r=b} d B=\frac{\mu_0 N I}{2(b-a)} \ln \frac{b}{a}
\end{aligned}
$$
$\therefore$ Correct answer is (a).
Analysis of Question
(i) If we see this problem independently, then I will rate this question moderately difficult. But the idea of this question is taken from question number $3.245$ of IE Irodov.
(ii) Interestingly the same question was asked in IIT-JEE 2001 also.
$$
d N=\frac{N}{b-a} d r
$$
Magnetic field due to this element at the centre of the coil will be
$$
\begin{aligned}
& d B=\frac{\mu_0(d N) I}{2 r}=\frac{\mu_0 N I}{2(b-a)} \frac{d r}{r} \\
\therefore \quad & B=\int_{r=a}^{r=b} d B=\frac{\mu_0 N I}{2(b-a)} \ln \frac{b}{a}
\end{aligned}
$$
$\therefore$ Correct answer is (a).
Analysis of Question
(i) If we see this problem independently, then I will rate this question moderately difficult. But the idea of this question is taken from question number $3.245$ of IE Irodov.
(ii) Interestingly the same question was asked in IIT-JEE 2001 also.
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