Question

A long solenoid is carrying a time dependent current such that the magnetic field inside has the form $\vec{B}(t)=B_0 t^2 \hat{k}$, where $\hat{k}$ is along the axis of the solenoid. The displacement current at the point $P$ on a circle of radius $r$ in a plane perpendicular to the axis

• A. is inversely proportional to $r$ and radially outward.
• B. is inversely proportional to $r$ and tangential.
• C. increases linearly with time and is tangential.
• D. is inversely proportional to $r^2$ and tangential.

Hint:

A time-varying axial magnetic field always produces a tangential electric field around the axis (circular symmetry).

Answer 1

The Correct Option is B

Step 1: Use Maxwell–Ampère law with displacement current.
$\displaystyle \nabla \times \vec{B} = \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}$.
Given $B(t) = B_0 t^2$, the time derivative is $\displaystyle \frac{\partial B}{\partial t} = 2 B_0 t$.

Step 2: Apply Faraday's law to circulating electric field.
The changing magnetic field induces a circulating (tangential) electric field around the axis: $\displaystyle \oint \vec{E}\cdot d\vec{l} = -\frac{d\Phi_B}{dt}$.
Since $\Phi_B = B_0 t^2 \cdot \pi r^2$, $\displaystyle \frac{d\Phi_B}{dt} = 2 B_0 t \, \pi r^2$.

Step 3: Find E-field direction and magnitude.
For circular symmetry: $\displaystyle E \cdot (2\pi r) = 2B_0 t\,\pi r^2$.
Thus $\displaystyle E = B_0 t\, r$.
Therefore, the displacement current density $\displaystyle J_d = \epsilon_0 \frac{\partial E}{\partial t} \propto t$ (linear in time).

Step 4: Direction of the field.
Induced electric field circulates around the solenoid ⇒ tangential direction.

Step 5: Conclusion.
The displacement current increases linearly with time and is tangential.