Physics - Magnetism Question with Solution | TestHub

PhysicsMagnetismMiscellaneousHard2 minPYQ_2020

PhysicsHardsingle choice

A disc of radius $R$ rotates with a constant angular velocity $ω$ about its own axis. The surface charge density of this disc varies as $σ = {ar}^{2}$ where $r$ is the distance from the centre of the disc. Determine the magnetic field intensity at the centre of the disc.

Options:

Answer:

Solution:

$dA = (2 π r) \cdot dr$

$dq = σ \cdot dA = (2 πa r^{3}) \cdot dr$

$i = (dq) f = (dq) \frac{ω}{2 π} = (aω r^{3}) \cdot dr$

$dB = \frac{μ_{0} i}{2 r} = \frac{{aμ}_{0} ω r^{2} dr}{2}$

$∴ B = \int_{0}^{R} dB = \frac{μ_{0} {aωR}^{3}}{6}$

Stream:JEESubject:PhysicsTopic:MagnetismSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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