Question

A cylindrical conductor of radius $R$ carrying current $i$ along the axis such that the magnetic field inside the conductor varies as $B=B_0{r}^2(0<r\le R)$, then which of the following is incorrect?

• A. Current density at distance ‘r’ from central axis of the conductor is $J(r)=\frac{2B_{0}r}{{\left(\mu \right)}_0}$
• B. Current density at distance ‘r’ from central axis of the conductor is $J(r)=\frac{3B_{0}r}{{\left(\mu \right)}_0}$
• C. Half of the total current would be confined within the radius of $\frac{R}{{\left(2\right)}^{\frac{1}{3}}}$ of the conductor
• D. For 4r > R$magneticfieldwouldvaryas$B(r)=\frac{B_{0} R^{3}}{r}$

Answer 1

Answer: (A)

Hint: Use Ampere’s law
Sol: As $B\left(r\right)=B_0{r}^2$



$\therefore$ Total current (upto $R)$ $B_0{R}^{2}2\pi R={\mu }_0{I}_0\Rightarrow I_0=\frac{2\pi R^3{B}_0}{{\mu }_0}$
And let upto 'r' it confines half of the current.
Then $B_0{r}^{\prime 2}\cdot 2\pi r^{\prime }={\mu }_0\frac{I_0}{2}$
$\Rightarrow B_0\cdot 2\pi r^{\prime 3}=\frac{{\mu }_0}{2}\cdot \frac{2\pi R^3{B}_0}{{\mu }_0}$
$\Rightarrow r^{\prime }=\frac{R}{(2{)}^{\frac{1}{3}}}$
Now let $\vec{J}(r)$ is the current density at ' $r$ Then $B_0{r}^2\cdot 2\pi r={\mu }_0{\int }_0^{r}J(r)2\pi rdr$
$\because \left(\int Bdl={\mu }_{0}I\right)$
$\Rightarrow 2\pi B_0{r}^3={\mu }_0\cdot 2\pi {\int }_0^{r}J(r)rdr$
$\therefore$ Differentiating both side w.r.t( $(r)$ $2\pi B_0\cdot 3r^2\cdot dr={\mu }_{0}2\pi \cdot J(r)rdr$
$\Rightarrow J(r)=\frac{3B_{0}r}{{\mu }_0}$
For $r>R$
$B\times 2\pi r={\mu }_0{I}_0$
$\Rightarrow B\times 2\pi r={\mu }_0\times \frac{2\pi R^3{B}_0}{{\mu }_0}$
$\Rightarrow B=\frac{B_0{R}^3}{r}$