JEE MAIN - Physics (2021 - 26th August Morning Shift - No. 1)
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)
$${3 \over 2}{{{a^2}} \over {{r^2}}}$$
$${2 \over 3}{{{a^2}} \over {{r^2}}}$$
$${2 \over 3}{{{r^2}} \over {{a^2}}}$$
$${3 \over 2}{{{r^2}} \over {{a^2}}}$$
Explanation
$${B_{axis}} = {{{\mu _0}i{R^2}} \over {2{{({R^2} + {x^2})}^{3/2}}}}$$
$${B_{centre}} = {{{\mu _0}i} \over {2R}}$$
$$\therefore$$ $${B_{centre}} = {{{\mu _0}i} \over {2a}}$$
$$\therefore$$ $${B_{axis}} = {{{\mu _0}i{a^2}} \over {2{{({a^2} + {r^2})}^{3/2}}}}$$
$$\therefore$$ fractional change in magnetic field =
$${{{{{\mu _0}i} \over {2a}} - {{{\mu _0}i{a^2}} \over {2{{({a^2} + {r^2})}^{3/2}}}}} \over {{{{\mu _0}i} \over {2a}}}} = 1 - {1 \over {{{\left[ {1 + \left( {{{{r^2}} \over {{a^2}}}} \right)} \right]}^{3/2}}}}$$
$$ \approx 1 - \left[ {1 - {3 \over 2}{{{r^2}} \over {{a^2}}}} \right] = {3 \over 2}{{{r^2}} \over {{a^2}}}$$
Note : $${\left( {1 + {{{r^2}} \over {{a^2}}}} \right)^{ - 3/2}} \approx \left( {1 - {3 \over 2}{{{r^2}} \over {{a^2}}}} \right)$$
[True only if r << a]
Hence, option (d) is the most suitable option.
$${B_{centre}} = {{{\mu _0}i} \over {2R}}$$
$$\therefore$$ $${B_{centre}} = {{{\mu _0}i} \over {2a}}$$
$$\therefore$$ $${B_{axis}} = {{{\mu _0}i{a^2}} \over {2{{({a^2} + {r^2})}^{3/2}}}}$$
$$\therefore$$ fractional change in magnetic field =
$${{{{{\mu _0}i} \over {2a}} - {{{\mu _0}i{a^2}} \over {2{{({a^2} + {r^2})}^{3/2}}}}} \over {{{{\mu _0}i} \over {2a}}}} = 1 - {1 \over {{{\left[ {1 + \left( {{{{r^2}} \over {{a^2}}}} \right)} \right]}^{3/2}}}}$$
$$ \approx 1 - \left[ {1 - {3 \over 2}{{{r^2}} \over {{a^2}}}} \right] = {3 \over 2}{{{r^2}} \over {{a^2}}}$$
Note : $${\left( {1 + {{{r^2}} \over {{a^2}}}} \right)^{ - 3/2}} \approx \left( {1 - {3 \over 2}{{{r^2}} \over {{a^2}}}} \right)$$
[True only if r << a]
Hence, option (d) is the most suitable option.
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