Question #c27e7

1 Answer
Jul 3, 2016

Answer:

(3)

Explanation:

Consider an element of thickness #dr# at a distance #r# from the centre of the coil.
The number of turns in this element are #dN=N/(b−a)dr#

Magnetic field* at the centre of the coil due to this element is given by the expression #dB=(μ_0dN.I)/(2r)#
Inserting values of #dN# we obtain
#dB=(μ_0N/(b−a)dr.I)/(2r)#
#=>dB=(μ_0NI)/(2(b−a)) 1/rdr#
Total magnetic field is found by Integrating from limits #a" to "b#

#B=int_a^b(μ_0NI)/(2(b−a)) 1/rdr#
#=>(μ_0NI)/(2(b−a))ln(|r|)|_a^b#, ignoring constant of integration as it is proper/definite integral
#=>(μ_0NI)/(2(b−a))(lnb-lna)#
#=>(μ_0NI)/(2(b−a))ln(b/a)#

-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-..-
*Magnetic field at Center of a coiled wire of radius #R# carrying a current #I# can be calculated as below:

Let #dvecl# be the infinitesimal length of the coil, along the circumference, carrying current and
#dhatr# be the unit vector along the radius of the coil.
The magnetic field #dvecB# produced by this element is given by the simplified expression for Biot-Savart Law
#dvecB=(mu_0Idvecl xxdhatr)/(4piR^2)#
which simplifies as the angle #=90^@# all elements along the path and the distance to the centre is constant. The scalar part becomes
#dB=(mu_0Icdotdl )/(4piR^2)#

To find the total magnetic field we need to find the line integral of the length which is the circumference #2piR#, due to symmetry. #B=(mu_0I)/(4piR^2) oint" "dl#
#=>B=(mu_0I)/(4piR^2) 2piR#
#=>B=(mu_0I)/(2R)#
If there are #N# number of turns, the total current is #NI# and magnetic field becomes
#B=(mu_0NI)/(2R)#