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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.

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.

Grade:12

2 Answers

Vikas TU
14149 Points
4 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
 
Kushagra Madhukar
askIITians Faculty 629 Points
11 months 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
 
Thanks and regrads,
Kushagra

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