Physics - Magnetism Question with Solution | TestHub

PhysicsMagnetismBiot-savart's & Ampere's Circuital lawMedium2 minPYQ_2021

PhysicsMediumsingle choice

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 $r < a$ )

Options:

Answer:

Solution:

As we know,

$B_{axis} = \frac{μ_{0} {iR}^{2}}{2 {(R^{2} + x^{2})}^{3 / 2}} where, R = a$

$B_{centre} = \frac{μ_{0} i}{2 a}$

$∴ B_{axis} = \frac{μ_{0} {ia}^{2}}{2 {(a^{2} + r^{2})}^{3 / 2}}$

$∴$ fractional change in magnetic field $= \frac{\frac{μ_{0} i}{2 a} - \frac{μ_{0} i a^{2}}{2 {(a^{2} + r^{2})}^{3 / 2}}}{\frac{μ_{0} i}{2 a}} = 1 - \frac{1}{{[1 + (\frac{r^{2}}{a^{2}})]}^{3 / 2}}$

$\approx 1 - [1 - \frac{3}{2} \frac{r^{2}}{a^{2}}] = \frac{3}{2} \frac{r^{2}}{a^{2}}$

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

⏱ 2mℹ️ Source: PYQ_2021

Doubts & Discussion

Loading discussions...