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JEE Mains · Physics · STD 12 - Moving charges and magnetism
  • JEE Mains 2018
  • Medium
  • MCQ

The dipole moment of a circular loop carrying a current \(I\), is \(m\) and the magnetic field at the centre of the loop is \(B_1\) . When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is \(B_2\) . The ratio \(\frac{{{B_1}}}{{{B_2}}}\) is

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Answer & Solution

Correct Answer

(B) \(\sqrt 2 \)

Step-by-step Solution

Detailed explanation

Magnetic field at the centre of \(\operatorname{loop}, B_{1}=\frac{\mu_{0} I}{2 R}\)

Dipole moment of circular loop is \(m=I A\)

\(\mathrm{m}_{1}=\mathrm{I} \cdot \mathrm{A}=\mathrm{I} \pi \mathrm{R}^{2} \quad\{\mathrm{R}=\text { Radius of the loop }\}\)

If moment is doubled (keeping current constant) \(R\) becomes \(\sqrt{2 \mathrm{R}}\)

\({{\text{m}}_2} = {\text{L}}\pi {(\sqrt 2 {\text{R}})^2} = 2.{\text{I}}\pi {{\text{R}}^2} = 2{{\text{m}}_1}\)

\(B_{2}=\frac{\mu_{0} I}{2(\sqrt{2} R)}\)

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