Question:
A plane electromagnetic wave travels in free space along \(z\)-axis. At a particular point in space, the electric field along \(x\)-axis is \(8.7 \, Vm^{-1}\). The magnetic field along \(y\)-axis is
A plane electromagnetic wave travels in free space along \(z\)-axis. At a particular point in space, the electric field along \(x\)-axis is \(8.7 \, Vm^{-1}\). The magnetic field along \(y\)-axis is
Show Hint
For an electromagnetic wave in free space, always use:
\[
E=cB
\]
where \(c=3\times10^8 \, m/s\).
Updated On: Jun 15, 2026
- \(2.9 \times 10^{-8} \, T\)
- \(3 \times 10^{-6} \, T\)
- \(8.7 \times 10^{-6} \, T\)
- \(3 \times 10^{-5} \, T\)
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The Correct Option is A
Solution and Explanation
Step 1: Use the relation between electric and magnetic fields in an electromagnetic wave.
For a plane electromagnetic wave travelling in free space, the magnitudes of electric field and magnetic field are related by
\[ \frac{E}{B}=c \] where \(c\) is the speed of light in free space.
\[ c=3\times10^8 \, m/s \]
Step 2: Write the given electric field.
The electric field is given as
\[ E=8.7 \, Vm^{-1} \]
Step 3: Calculate the magnetic field.
From the relation,
\[ B=\frac{E}{c} \] Substituting values,
\[ B=\frac{8.7}{3\times10^8} \] \[ B=2.9\times10^{-8} \, T \]
Step 4: Check the direction of fields.
The wave travels along the \(z\)-axis.
The electric field is along the \(x\)-axis and the magnetic field is along the \(y\)-axis, so \(\vec{E}\), \(\vec{B}\), and direction of propagation are mutually perpendicular.
Step 5: Final conclusion.
Hence, the magnetic field is
\[ \boxed{2.9\times10^{-8} \, T} \]
For a plane electromagnetic wave travelling in free space, the magnitudes of electric field and magnetic field are related by
\[ \frac{E}{B}=c \] where \(c\) is the speed of light in free space.
\[ c=3\times10^8 \, m/s \]
Step 2: Write the given electric field.
The electric field is given as
\[ E=8.7 \, Vm^{-1} \]
Step 3: Calculate the magnetic field.
From the relation,
\[ B=\frac{E}{c} \] Substituting values,
\[ B=\frac{8.7}{3\times10^8} \] \[ B=2.9\times10^{-8} \, T \]
Step 4: Check the direction of fields.
The wave travels along the \(z\)-axis.
The electric field is along the \(x\)-axis and the magnetic field is along the \(y\)-axis, so \(\vec{E}\), \(\vec{B}\), and direction of propagation are mutually perpendicular.
Step 5: Final conclusion.
Hence, the magnetic field is
\[ \boxed{2.9\times10^{-8} \, T} \]
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