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 × 10-9 T, then the charge carried by the ring is close to (μ0 = 4 π × 10-7 N/A2) :

Question

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 × 10-9 T, then the charge carried by the ring is close to (μ0 = 4 π × 10-7 N/A2) : (A)  2 × 10-6 C (B)  3 × 10-5 C (C) 4 × 10-5 C (D)  7 × 10-6 C

Tardigrade Admin · Accepted Answer

B =(μ0 i/2R) = (μ0qω/2R 2π) ⇒ q=3 ×10-5 C

$2 \times 1 0^{- 6} C$

$3 \times 1 0^{- 5} C$

$4 \times 1 0^{- 5} C$

$7 \times 1 0^{- 6} C$

$B = \frac{μ _{0} i}{2 R} = \frac{μ _{0} q ω}{2 R 2 π}$
$\Rightarrow q = 3 \times 1 0^{- 5} C$

2×10−6C

3×10−5C

4×10−5C

7×10−6C

B=2Rμ0​i​=2R2πμ0​qω​ ⇒q=3×10−5C

$2 \times 1 0^{- 6} C$

$3 \times 1 0^{- 5} C$

$4 \times 1 0^{- 5} C$

$7 \times 1 0^{- 6} C$

$B = \frac{μ _{0} i}{2 R} = \frac{μ _{0} q ω}{2 R 2 π}$
$\Rightarrow q = 3 \times 1 0^{- 5} C$