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

• A. Current density at distance ‘r’ from central axis of the conductor is $J(r) \, =\frac{2 B_{0} r}{\left(\mu \right)_{0}}$
• B. Current density at distance ‘r’ from central axis of the conductor is $J(r) \, =\frac{3 B_{0} r}{\left(\mu \right)_{0}}$
• 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$ magnetic field would vary as $B(r)=\frac{B_{0} R^{3}}{r}$

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 $B_{0} r^{2} \cdot 2 \pi r=\mu_{0} \int_{0}^{r} J(r) 2 \pi r d r$
$\because\left(\int B d l=\mu_{0} I\right)$
$\Rightarrow 2 \pi B_{0} r^{3}=\mu_{0} \cdot 2 \pi \int_{0}^{r} J(r) r d r$
$\therefore $ Differentiating both side w.r.t( $(r)$ $2 \pi B_{0} \cdot 3 r^{2} \cdot d r=\mu_{0} 2 \pi \cdot J(r) r d r$
$\Rightarrow J(r)=\frac{3 B_{0} r}{\mu_{0}}$
For $r>R$
$B \times 2 \pi r=\mu_{0} I_{0}$
$\Rightarrow B \times 2 \pi r=\mu_{0} \times \frac{2 \pi R^{3} B_{0}}{\mu_{0}}$
$\Rightarrow B=\frac{B_{0} R^{3}}{r}$