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A circular coil carrying current ‘I’ has a radius ‘R’ and magnetic field at the center is ‘B’. At what distance from the center along the axis of the same coil, the magnetic field will be B8?
A circular coil carrying current ‘I’ has a radius ‘R’ and magnetic field at the center is ‘B’. At what distance from the center along the axis of the same coil, the magnetic field will be B8?
Physics | Physics
A circular coil carrying current ‘I’ has a radius ‘R’ and magnetic field at the center is ‘B’. At what distance from the center along the axis of the same coil, the magnetic field will be B8?

  1. R√3
  2. 2R
  3. 3R
  4. R√2

Solution: The correct answer is A. R√3

We know that the magnetic field B is given by at a distance x from the center of a current-carrying coil’s axis.

$ {{B}_{1}}=\frac{{{\mu }_{0}}I{{R}^{2}}}{2{{\left( {{R}^{2}}+{{x}^{2}} \right)}^{3/2}}}\,and\,B=\frac{{{\mu }_{0}}I}{2R} $

$ {{B}_{1}}=\frac{B}{8} $

$ \frac{{{\mu }_{0}}I{{R}^{2}}}{2{{\left( {{R}^{2}}+{{x}^{2}} \right)}^{3/2}}}=\frac{{{\mu }_{0}}I}{16R} $

$ 8{{R}^{3}}={{\left( {{R}^{2}}+{{x}^{2}} \right)}^{3/2}} $

$ 4{{R}^{2}}={{R}^{2}}+{{x}^{2}} $

$ {{x}^{2}}=3{{R}^{2}} $

$ x=R\sqrt{3} $

 

 

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