Question

Which of the following options is correct for transformation of electric field \(\vec{E}\) and magnetic field \(\vec{B}\) under time reversal, i.e., \(t \to -t\)?

• A. \(\vec{E} \to \vec{E}\) and \(\vec{B} \to \vec{B}\)
• B. \(\vec{E} \to -\vec{E}\) and \(\vec{B} \to \vec{B}\)
• C. \(\vec{E} \to \vec{E}\) and \(\vec{B} \to -\vec{B}\)
• D. \(\vec{E} \to -\vec{E}\) and \(\vec{B} \to -\vec{B}\)

Hint:

Charge density does not change under \(t\to-t\), but current density flips sign since velocity flips; \(\vec{E}\) follows charge, \(\vec{B}\) follows current.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Time reversal means replacing \(t\) by \(-t\) and asking how each physical quantity must change so the laws of physics still hold. Electric and magnetic fields are not fundamental on their own, they come from charges and currents, so first work out how charge density and current density behave under \(t\to-t\), then use Maxwell's equations to see how \(\vec{E}\) and \(\vec{B}\) must behave.

Step 2: Key Formula or Approach:
1. Charge density \(\rho\) counts charges sitting at fixed positions, so it does not depend on which way time runs: \(\rho \to \rho\) (even).
2. Current density is \(\vec{J} = \rho\vec{v}\), with \(\vec{v}\) a velocity. Under \(t\to-t\), a trajectory \(\vec{r}(t)\) traces the same path backward, so velocity \(\vec{v} = d\vec{r}/dt\) flips sign: \(\vec{v}\to-\vec{v}\), hence \(\vec{J}\to-\vec{J}\) (odd).
3. \(\vec{E}\) is sourced by \(\rho\) through Coulomb's law and Gauss's law, so it transforms the same way as \(\rho\). \(\vec{B}\) is sourced by \(\vec{J}\) through the Biot-Savart law and Ampere's law, so it transforms the same way as \(\vec{J}\).

Step 3: Detailed Explanation:
Since \(\rho\to\rho\) is unchanged, and \(\vec{E}\) is generated directly by charge distributions through \(\vec{E} \propto \rho\), it follows that:
\[ \vec{E} \to \vec{E} \]
Since \(\vec{J}\to-\vec{J}\), and \(\vec{B}\) is generated by currents through the Biot-Savart law \(\vec{B}\propto \vec{J}\times \hat{r}\,/r^2\) integrated over the source, it follows that:
\[ \vec{B} \to -\vec{B} \]
This also matches the everyday picture of a bar magnet, whose field comes from circulating electron currents. Reversing the direction of time reverses the sense of every circulating charge, which reverses the current loop's field.

Step 4: Why the other options are wrong.
Option (A) keeps both fields unchanged, missing that current, and so \(\vec{B}\), depends on velocity and must flip. Option (B) flips \(\vec{E}\) instead of \(\vec{B}\), which would wrongly make a static charge distribution produce a different field under time reversal even though \(\rho\) itself does not change. Option (D) flips both fields, over-correcting by treating \(\vec{E}\) as if it depended on velocity the way \(\vec{B}\) does.

Final Answer:
\(\vec{E}\) is even and \(\vec{B}\) is odd under time reversal, because \(\vec{E}\) traces charge, which does not change with the direction of time, while \(\vec{B}\) traces current, which reverses with velocity. \[ \boxed{\vec{E}\to\vec{E},\ \vec{B}\to-\vec{B}} \]