# two parallel coaxial circular coils of equal radius R and equal no of turns N carry equal currents I in the same direction and are represented by a distance 2R. find the magnitude and direction of the net magnetic field produced at the mid-point of the line joining their centres.

Vikas TU
14149 Points
6 years ago
. Radius of 2 parallel co-axial circular coil = R
Number of turns = N
Current in both coils = I
Distance between the 2 coils = 2R

As the point where we have to find the magnetic field is half way between the 2 coils, it will be at a distance of R from center of both the coils.

So the magnetic field at the point because of coil 1 = B1

So, B1 = (μoNIR2) / 2[(R+R)2+R]3/2

Similarly, B2 = (μoNIR2) / 2[(R+R)2+R]3/2

So, the total magnetic field, B = B1 + B2

• B = B1 + B2
= (μoNIR2) / 2[(R+R)2+R]3/2  + (μoNIR2) / 2[(R+R)2+R]3/2
= (μoNIR2) / [(R+R)2+R]3/2
= (μoNIR2) / [4R2+R] 3/2
= (μoNIR2) /8R3[1+1/R]3/2
Since, R>>1 we can say
B = (μoNIR2) /8R3

askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your problem.

Radius of 2 parallel co-axial circular coil = R
Number of turns = N
Current in both coils = I
Distance between the 2 coils = 2R
As the point where we have to find the magnetic field is half way between the 2 coils, it will be at a distance of R from center of both the coils.
So the magnetic field at the point because of coil 1 = B1
So, B1 = (μoNIR2) / 2[(R+R)2+R]3/2
Similarly, B2 = (μoNIR2) / 2[(R+R)2+R]3/2
So, the total magnetic field, B = B1 + B2
B = B1 + B2  = (μoNIR2) / 2[(R+R)2+R]3/2  + (μoNIR2) / 2[(R+R)2+R]3/2
= (μoNIR2) / [(R+R)2+R]3/2
= (μoNIR2) / [4R2+R] 3/2
= (μoNIR2) /8R3[1+1/R]3/2
Since, R>>1 we can say
B = (μoNIR2) /8R3