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Question: A thin insulated wire forms a plane spiral of \(N = 100\) tight turns carrying a current \(i = 8\,mA...

A thin insulated wire forms a plane spiral of N=100N = 100 tight turns carrying a current i=8mAi = 8\,mA. The radii of inside and outside turns are equal to a=50mma = 50\,mm and b=100mmb = 100\,mm. Find
(a) The magnetic induction at the centre of the spiral
(b) The magnetic moment of the spiral with given current

Explanation

Solution

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:
Bo{B_o} = μI2r\dfrac{{\mu I}}{{2r}}
The radius of the circular loop varies from aa to bb. Therefore, total magnetic induction at the centre is:
Br{B_r} = μI2rdN\smallint \dfrac{{\mu I}}{{2r}}dN....................(1)
(where μI2r\dfrac{{\mu I}}{{2r}} is magnetic induction due to one turn of radius rr and dNdN is the number of turns in the interval (rr, r+drr+dr)i.e.
dN=NbadrdN =\dfrac{N}{{b - a}}dr
Substituting value of dN in eq (1) and then integrating between a and b, we obtain
Bo{B_o} = abμI2rNbadr\int_a^b {\dfrac{{\mu I}}{{2r}}} \dfrac{N}{{b - a}}dr
Bo\Rightarrow {B_o}= μIN2(ba)lnba\dfrac{{\mu IN}}{{2(b - a)}}\ln \dfrac{b}{a}
Bo\Rightarrow {B_o} = 4π×107×100×8×1032(50×103)×2.303\dfrac{{4\pi \times {{10}^{ - 7}} \times 100 \times 8 \times {{10}^{ - 3}}}}{{2(50 \times {{10}^{ - 3}})}} \times 2.303
Bo\therefore {B_o}= 7μT7\mu T

(b) Magnetic moment of a turn of radius rr is
dM=Ndrba×iπr2dM =\dfrac{{Ndr}}{{b - a}} \times i\pi {r^2}
Total magnetic moment of all turns is
M=dMM = \int {dM} (1)
Substituting value of dM in eq(1), we get
M=Nbaiπb3a33M = \dfrac{N}{{b - a}}i\pi \dfrac{{{b^3} - {a^3}}}{3}
M=100(10050)×103×8×1034π×107(0.130.0533)\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})
M=15mA\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.