A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of $40 \pi \mathrm{rad} \mathrm{s}^{-1}$ about its axis, perpendicular to its plane. If the magnetic field at its centre is $3.8 \times 10^{-9} \mathrm{~T}$, then the charge carried by the ring is close to ( $\mu_{0}=4 \pi \times 10^{-7} \mathrm{~N} / \mathrm{A}^{2}$ ).

(D) $3 \times 10^{-5} \mathrm{C}$

(A) $2 \times 10^{-6} \mathrm{C}$

(B) $7 \times 10^{-6} \mathrm{C}$

(C) $4 \times 10^{-5} \mathrm{C}$

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Magnetic Effects of Current

A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of

40 π rad s^{- 1}

about its axis, perpendicular to its plane. If the magnetic field at its centre is

3.8 \times 1 0^{- 9} T

, then the charge carried by the ring is close to (

μ_{0} = 4 π \times 1 0^{- 7} N / A^{2}

Answer: (D)

A ring carrying charge

q

and spinning at angular speed

ω

behaves like a current loop of current

i = \frac{q ω}{2 π}

, so the field at its centre is

B = \frac{μ _{0} i}{2 R} = \frac{μ _{0} q ω}{4 π R}

. Solving for the charge,

q = \frac{4 π R B}{μ _{0} ω} = \frac{4 π \times 0.1 \times 3.8 \times 1 0 ^{- 9}}{4 π \times 1 0 ^{- 7} \times 40 π} \approx 3 \times 1 0^{- 5} C

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option