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Current Loop as a Magnetic Dipole question from JEE Main 2018

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Current Loop as a Magnetic Dipole

The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is B1. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is B2B_2. The ratio B1B2\frac{B_1}{B_2} is:

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Answer: (c)

  1. m=IπR2m = I\pi R^2; doubling mm at the same II gives R2=2R1R_2 = \sqrt{2}R_1.
  2. B=μ0I2R∝1RB = \frac{\mu_0 I}{2R} \propto \frac{1}{R}
  3. B1B2=R2R1=2\frac{B_1}{B_2} = \frac{R_2}{R_1} = \sqrt{2}, so the answer is (c).
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