Medium MCQ +4 / -1 PYQ · JEE Mains 2021

The fractional change in the magnetic field intensity at a distance 'r' from centre on the axis of current carrying coil of radius 'a' to the magnetic field intensity at the centre of the same coil is : (Take r < a)

  1. A ${3 \over 2}{{{a^2}} \over {{r^2}}}$
  2. B ${2 \over 3}{{{a^2}} \over {{r^2}}}$
  3. C ${2 \over 3}{{{r^2}} \over {{a^2}}}$
  4. D ${3 \over 2}{{{r^2}} \over {{a^2}}}$ Correct answer

Solution

${B_{axis}} = {{{\mu _0}i{R^2}} \over {2{{({R^2} + {x^2})}^{3/2}}}}$<br><br>${B_{centre}} = {{{\mu _0}i} \over {2R}}$<br><br>$\therefore$ ${B_{centre}} = {{{\mu _0}i} \over {2a}}$<br><br>$\therefore$ ${B_{axis}} = {{{\mu _0}i{a^2}} \over {2{{({a^2} + {r^2})}^{3/2}}}}$<br><br>$\therefore$ fractional change in magnetic field =<br><br>$${{{{{\mu _0}i} \over {2a}} - {{{\mu _0}i{a^2}} \over {2{{({a^2} + {r^2})}^{3/2}}}}} \over {{{{\mu _0}i} \over {2a}}}} = 1 - {1 \over {{{\left[ {1 + \left( {{{{r^2}} \over {{a^2}}}} \right)} \right]}^{3/2}}}}$$<br><br>$$ \approx 1 - \left[ {1 - {3 \over 2}{{{r^2}} \over {{a^2}}}} \right] = {3 \over 2}{{{r^2}} \over {{a^2}}}$$<br><br>Note : $${\left( {1 + {{{r^2}} \over {{a^2}}}} \right)^{ - 3/2}} \approx \left( {1 - {3 \over 2}{{{r^2}} \over {{a^2}}}} \right)$$<br><br>[True only if r &lt;&lt; a]<br><br>Hence, option (d) is the most suitable option.

About this question

Subject: Physics · Chapter: Magnetic Effects of Current · Topic: Biot-Savart Law

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