Surface charge density on a ring of radius a | vedclass

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Question

(A) The surface charge density is $\sigma$.The total charge $q$ on the ring is given by the product of surface charge density,circumference,and width:

Vedclass · Accepted Answer

$\pi \mu_0 f \sigma d$

Vedclass · Answer

(A) The surface charge density is $\sigma$.The total charge $q$ on the ring is given by the product of surface charge density,circumference,and width:$q = \sigma (2 \pi a) d$The equivalent current $i$ produced by the rotation of the ring with frequency $f$ is:$i = \frac{q}{T} = qf = \sigma (2 \pi a) d f$The magnetic field $B$ at the centre of a circular current-carrying loop is given by:$B = \frac{\mu_0 i}{2a}$Substituting the value of $i$ into the formula:$B = \frac{\mu_0 (\sigma 2 \pi a d f)}{2a}$Simplifying the expression:$B = \pi \mu_0 \sigma d f$Thus,the correct option is $A$.

Surface charge density on a ring of radius $a$ and width $d$ is $\sigma$ as shown in the figure. It rotates with frequency $f$ about its own axis. Assume that the charge is only on the outer surface. The magnetic field induction at the centre is (Assume that $d \ll a$):

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