Magnetics Ques 57
- A thin ring of $10 cm$ radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of $40 \pi rad s^{-1}$ about its axis, perpendicular to its plane. If the magnetic field at its centre is $3.8 \times 10^{-9} T$, then the charge carried by the ring is close to $\left(\mu _0=4 \pi \times 10^{-7} N / A^{2}\right)$.
(a) $2 \times 10^{-6} C$
(b) $3 \times 10^{-5} C$
(c) $4 \times 10^{-5} C$
(d) $7 \times 10^{-6} C$
(2019 Main, 12 April I)
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Answer:
Correct Answer: 57.(b)
Solution:
Formula:
Magnetic Field Due To A Straight Wire:
- Given, $\mu _0=4 \pi \times 10^{-7} N / A^{2}$,
$$ \begin{aligned} & \omega & =40 \pi rad / s \\ B _{\text {at centre }} & & =3.8 \times 10^{-9} T \\ \text { and } & R & =10 cm=0.1 m \end{aligned} $$
Now, we know that, magnetic field at the centre of a current carrying ring is given by
$$ B=\frac{\mu _0 I}{2 r} $$
Here, $I$ can be determined by flow of charge per rotation, i.e.
$$ \begin{aligned} & I=\frac{Q}{T} \\ & \text { Here, } \quad T=\frac{2 \pi}{\omega} \\ & \Rightarrow \quad I=\frac{Q \omega}{2 \pi} \end{aligned} $$
By putting value of $I$ from Eq. (iii) to Eq. (i), we get
$$ \begin{aligned} B & =\frac{\mu _0 Q \omega}{2 r \times 2 \pi} \text { or } Q=\frac{2 B r \times 2 \pi}{\mu _0 \omega} \\ & =\frac{2 \times 3.8 \times 10^{-9} \times 0.1 \times 2 \pi}{4 \pi \times 10^{-7} \times 40 \pi} \\ & =\frac{2 \times 3.8 \times 0.1}{2 \times 40 \pi} \times 10^{-2} \\ & =0.003022 \times 10^{-2} C \\ & =3.022 \times 10^{-5} C \\ \text { or } \quad Q & =3 \times 10^{-5} C \end{aligned} $$
BETA
