A non-conducting disc of radius 2 m is rotated about its axis of symmetry perpendicular to the plane of disc with a uniform angular speed ω. The disc carries a charge whose density is given as σ=A-B r, where A and B are positive constants and r is distance from the centre of disc. The value of A / B for which magnetic field at the centre of disc will be zero is

Question

A non-conducting disc of radius 2 m is rotated about its axis of symmetry perpendicular to the plane of disc with a uniform angular speed ω. The disc carries a charge whose density is given as σ=A-B r, where A and B are positive constants and r is distance from the centre of disc. The value of A / B for which magnetic field at the centre of disc will be zero is (A) 1 (B) 2 (C) (2/3) (D) (4/3)

Tardigrade Admin · Accepted Answer

d i=(d q/((2 π)ω))=(ω/2 π)(σ .2 π r d r)=ω(A-B r) r d r d B=(μ0 ω(A-B r) r d r/2 r) ; B=(μ0 ω/2) ∫ limits02(A-B r) d r 0=[A r-B (r2/2)]02 2 A-2 B=0, (A/B)=1

1

2

$\frac{2}{3}$

$\frac{4}{3}$

$d i = \frac{d q}{( \frac{2 π}{ω} )} = \frac{ω}{2 π} (σ .2 π r d r) = ω (A - B r) r d r$
$d B = \frac{μ _{0} ω ( A - B r ) r d r}{2 r}; B = \frac{μ _{0} ω}{2} 0 \int 2 (A - B r) d r$
$0 = [A r - B \frac{r ^{2}}{2}]_{0}^{2}$
$2 A - 2 B = 0, \frac{A}{B} = 1$

1

2

32​

34​

di=(ω2π​)dq​=2πω​(σ.2πrdr)=ω(A−Br)rdr dB=2rμ0​ω(A−Br)rdr​;B=2μ0​ω​0∫2​(A−Br)dr 0=[Ar−B2r2​]02​ 2A−2B=0,BA​=1

$\frac{2}{3}$

$\frac{4}{3}$

$d i = \frac{d q}{( \frac{2 π}{ω} )} = \frac{ω}{2 π} (σ .2 π r d r) = ω (A - B r) r d r$
$d B = \frac{μ _{0} ω ( A - B r ) r d r}{2 r}; B = \frac{μ _{0} ω}{2} 0 \int 2 (A - B r) d r$
$0 = [A r - B \frac{r ^{2}}{2}]_{0}^{2}$
$2 A - 2 B = 0, \frac{A}{B} = 1$