
A thin insulated wire forms a plane spiral of $N = 100$ tight turns carrying a current $i = 8\,mA$. The radii of inside and outside turns are equal to $a = 50\,mm$ and $b = 100\,mm$. Find
(a) The magnetic induction at the centre of the spiral
(b) The magnetic moment of the spiral with given current
Answer
429k+ views
Hint:In order to solve this question, you must be aware of the concept of Biot-Savart’s law which describes the magnetic field generated by a constant electric current.The Biot Savart Law is an equation describing the magnetic field generated by a constant electric current.
Complete step by step answer:
(a) From Biot-Savart’s law, the magnetic induction due to a circular current carrying wire loop at its centre is given by:
${B_o}$ = $\dfrac{{\mu I}}{{2r}}$
The radius of the circular loop varies from $a$ to $b$. Therefore, total magnetic induction at the centre is:
${B_r}$ = $\smallint \dfrac{{\mu I}}{{2r}}dN$....................(1)
(where $\dfrac{{\mu I}}{{2r}}$ is magnetic induction due to one turn of radius $r$ and $dN$ is the number of turns in the interval ($r$, $r+dr$)i.e.
$dN =\dfrac{N}{{b - a}}dr $
Substituting value of dN in eq (1) and then integrating between a and b, we obtain
${B_o}$ = $\int_a^b {\dfrac{{\mu I}}{{2r}}} \dfrac{N}{{b - a}}dr$
$\Rightarrow {B_o}$= $\dfrac{{\mu IN}}{{2(b - a)}}\ln \dfrac{b}{a}$
$\Rightarrow {B_o}$ = $\dfrac{{4\pi \times {{10}^{ - 7}} \times 100 \times 8 \times {{10}^{ - 3}}}}{{2(50 \times {{10}^{ - 3}})}} \times 2.303$
$\therefore {B_o}$= $7\mu T$
(b) Magnetic moment of a turn of radius $r$ is
$dM =\dfrac{{Ndr}}{{b - a}} \times i\pi {r^2}$
Total magnetic moment of all turns is
$M = \int {dM} $ (1)
Substituting value of dM in eq(1), we get
$M = \dfrac{N}{{b - a}}i\pi \dfrac{{{b^3} - {a^3}}}{3}$
$\Rightarrow M = \dfrac{{100}}{{(100 - 50) \times {{10}^{ - 3}}}} \times 8 \times {10^{ - 3}}4\pi \times {10^{ - 7}}(\dfrac{{{{0.1}^3} - {{0.05}^3}}}{3})$
$\therefore M =15\,mA$
Note:Biot-Savart’s law is applicable for very small conductors which carry current. It is an equation that gives the magnetic field produced due to a current carrying segment.It relates the magnetic field to the magnitude, direction, length, and proximity of the electric current. Biot–Savart law is consistent with both Ampere’s circuital law and Gauss’s theorem.
Complete step by step answer:
(a) From Biot-Savart’s law, the magnetic induction due to a circular current carrying wire loop at its centre is given by:
${B_o}$ = $\dfrac{{\mu I}}{{2r}}$
The radius of the circular loop varies from $a$ to $b$. Therefore, total magnetic induction at the centre is:
${B_r}$ = $\smallint \dfrac{{\mu I}}{{2r}}dN$....................(1)
(where $\dfrac{{\mu I}}{{2r}}$ is magnetic induction due to one turn of radius $r$ and $dN$ is the number of turns in the interval ($r$, $r+dr$)i.e.
$dN =\dfrac{N}{{b - a}}dr $
Substituting value of dN in eq (1) and then integrating between a and b, we obtain
${B_o}$ = $\int_a^b {\dfrac{{\mu I}}{{2r}}} \dfrac{N}{{b - a}}dr$
$\Rightarrow {B_o}$= $\dfrac{{\mu IN}}{{2(b - a)}}\ln \dfrac{b}{a}$
$\Rightarrow {B_o}$ = $\dfrac{{4\pi \times {{10}^{ - 7}} \times 100 \times 8 \times {{10}^{ - 3}}}}{{2(50 \times {{10}^{ - 3}})}} \times 2.303$
$\therefore {B_o}$= $7\mu T$
(b) Magnetic moment of a turn of radius $r$ is
$dM =\dfrac{{Ndr}}{{b - a}} \times i\pi {r^2}$
Total magnetic moment of all turns is
$M = \int {dM} $ (1)
Substituting value of dM in eq(1), we get
$M = \dfrac{N}{{b - a}}i\pi \dfrac{{{b^3} - {a^3}}}{3}$
$\Rightarrow M = \dfrac{{100}}{{(100 - 50) \times {{10}^{ - 3}}}} \times 8 \times {10^{ - 3}}4\pi \times {10^{ - 7}}(\dfrac{{{{0.1}^3} - {{0.05}^3}}}{3})$
$\therefore M =15\,mA$
Note:Biot-Savart’s law is applicable for very small conductors which carry current. It is an equation that gives the magnetic field produced due to a current carrying segment.It relates the magnetic field to the magnitude, direction, length, and proximity of the electric current. Biot–Savart law is consistent with both Ampere’s circuital law and Gauss’s theorem.
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