Questions

A non – conducting ring of radius R having uniformly distributed charge Q starts rotating about $X X'$ axis passing through the centre in the plane of the ring with an angular acceleration $' α'$ as shown in the figure. Another small conducting ring having radius $a (a < < R)$ is kept fixed at the centre of bigger ring in such a way that the axis is passing through its centre and perpendicular to its plane. If the resistance of the small ring is $r = 1 Ω$ , find the induced current in it (in ampere).

(Given $Q = \frac{16 \times 10^{2}}{μ_{o}} C, R = 1 m, a = 0.1 m, α = 8 rad / s^{- 2}$ )

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detailed solution

Correct option is P

$d q = \frac{q d θ}{2 π}, d i = \frac{d q}{T} = \frac{q ω d θ}{2 π \cdot 2 π} = \frac{q ω}{4 π^{2}} d θ$

Magnetic Field on axis at center is given by,

$d B = \frac{μ_{O} d i (r)^{2}}{2 {(r^{2} + x^{2})}^{3 / 2}}$

here $r=Rsinθ$ and $\sqrt{r^{2} + x^{2}} = R$
$d B = \frac{μ_{O} d i (R \sin θ)^{2}}{2 R^{3}} \Rightarrow \int d B = \int_{o}^{2 π} \frac{μ_{O} \sin^{2} θ}{2 R} (\frac{q ω}{4 π^{2}}) d θ$

$\Rightarrow B = \frac{μ_{o} q ω}{8 π R} \emptyset = B π a^{2} = π a^{2} \cdot \frac{μ_{o} q ω}{8 π R} = \frac{μ_{o} q ω a^{2}}{8 R}$
$induced emf; \in= \frac{d \emptyset}{d t} = \frac{μ_{o} q a^{2}}{8 R} α = 16 V ∴ i = \frac{16}{1} = 16 A$

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detailed solution

Correct answer is 16

$d q = \frac{q d θ}{2 π}, d i = \frac{d q}{T} = \frac{q ω d θ}{2 π \cdot 2 π} = \frac{q ω}{4 π^{2}} d θ$

Magnetic Field on axis at center is given by,

$d B = \frac{μ_{O} d i (r)^{2}}{2 {(r^{2} + x^{2})}^{3 / 2}}$

here $r=Rsinθ$ and $\sqrt{r^{2} + x^{2}} = R$
$d B = \frac{μ_{O} d i (R \sin θ)^{2}}{2 R^{3}} \Rightarrow \int d B = \int_{o}^{2 π} \frac{μ_{O} \sin^{2} θ}{2 R} (\frac{q ω}{4 π^{2}}) d θ$