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: Use the Lorentz force law instead of the source picture.
A different route to the same conclusion is to demand that Newton's second law and the Lorentz force stay consistent when $t$ is replaced by $-t$. The force on a charge is $\vec{F} = q(\vec{E}+\vec{v}\times\vec{B})$, and separately $\vec{F}=m\vec{a}$.

Step 2: Work out how position, velocity and acceleration behave under $t\to-t$.
If $\vec{r}(t)$ is the trajectory, the time-reversed trajectory is $\vec{r}(-t)$, tracing the same path backward. Velocity is one time derivative, $\vec{v}=d\vec{r}/dt$, and one derivative with respect to $-t$ brings a minus sign, so $\vec{v}\to-\vec{v}$. Acceleration is two time derivatives, $\vec{a}=d^2\vec{r}/dt^2$, and the two minus signs cancel, so $\vec{a}\to\vec{a}$.

Step 3: The force must stay unchanged.
Mass $m$ does not change, and $\vec{a}\to\vec{a}$, so from $\vec{F}=m\vec{a}$ the force itself must stay unchanged: $\vec{F}\to\vec{F}$.

Step 4: Demand the Lorentz force formula gives back the same $\vec{F}$.
Let $\vec{E}\to\vec{E}'$ and $\vec{B}\to\vec{B}'$ be the transformed fields, with $\vec{v}\to-\vec{v}$ as found above. The transformed force is:
\[ \vec{F}' = q\left(\vec{E}' + (-\vec{v})\times\vec{B}'\right) = q\vec{E}' - q\,\vec{v}\times\vec{B}' \]
This must equal the original force $\vec{F} = q\vec{E}+q\,\vec{v}\times\vec{B}$ for every possible velocity the charge could have, since the field transformation cannot depend on one particular test charge's motion.

Step 5: Match the velocity-independent and velocity-dependent parts separately.
Since $\vec{v}$ is arbitrary, the part with no $\vec{v}$ and the part linear in $\vec{v}$ must match on their own:
\[ q\vec{E}' = q\vec{E} \implies \vec{E}' = \vec{E} \]
\[ -q\,\vec{v}\times\vec{B}' = q\,\vec{v}\times\vec{B} \implies \vec{B}' = -\vec{B} \]

Step 6: Compare with the options.
This gives $\vec{E}\to\vec{E}$ and $\vec{B}\to-\vec{B}$, which is option (C). Options (A), (B) and (D) all fail this consistency check, for instance option (B) would need $q\vec{v}\times\vec{B}$ to flip sign on both sides at once, which does not happen if $\vec{B}$ is left unchanged.

Final Answer:
Demanding that $\vec{F}=m\vec{a}$ stay consistent under time reversal forces $\vec{E}$ to stay the same and $\vec{B}$ to flip sign. \[ \boxed{\vec{E}\to\vec{E},\ \vec{B}\to-\vec{B}} \]