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Physics - Magnetism Question with Solution | TestHub

PhysicsMagnetismBiot-savart's & Ampere's Circuital lawEasy2 minPYQ_2018

PhysicsEasysingle choice

A circular coil carrying current $I$ has a radius $R$ and magnetic field at the centre is $B$ . The distance from the centre along the axis of the same coil where the magnetic field will be $\frac{B}{8}$ is

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Answer:

Solution:

$B_{center} = \frac{μ_{0} nI}{2 R} = B$
$B_{x} = \frac{B}{8} = \frac{μ_{0} {nIR}^{2}}{2 {(R^{2} + x^{2})}^{3 / 2}}$
$\frac{1}{8} (\frac{μ_{0} nI}{2 R}) = \frac{μ_{0} {nIR}^{2}}{2 {(R^{2} + x^{2})}^{3 / 2}}$
$\frac{1}{8 R^{3}} = \frac{1}{{(R^{2} + x^{2})}^{3 / 2}}$
$\frac{1}{2 R} = \frac{1}{{(R^{2} + x^{2})}^{\frac{1}{2}}}$
$4 R^{2} = R^{2} + x^{2}$
$x^{2} = 3 R^{2}$
$x = \sqrt{3} R$

Stream:MHT_CETSubject:PhysicsTopic:MagnetismSubtopic:Biot-savart's & Ampere's Circuital law

⏱ 2mℹ️ Source: PYQ_2018

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