Physics Q43 – JEE-MAIN 2026

An electromagnetic wave travelling in x-direction is described by field equation

E_{y} = 300 \sin ω (t - \frac{x}{c})

. If the electron is restricted to move in y-direction only with speed of

1.5 \times 10^{6}

m/s then ratio of maximum electric and magnetic forces acting on the electron is ______.

Clue 1 of 3: Understand the forces

The problem involves an electron moving in an electromagnetic wave, so both electric and magnetic forces will act on it.

🧠 Full Solution Path

Step 1: Identify maximum electric and magnetic field strengths✦ Active

The given electric field equation $E_{y} = 300 \sin ω (t - \frac{x}{c})$ indicates that the maximum electric field strength is $E_{0} = 300$ V/m. For an electromagnetic wave, the maximum magnetic field strength $B_{0}$ is related to $E_{0}$ by the equation $E_{0} = c B_{0}$ , where $c$ is the speed of light ( $3 \times 10^{8}$ m/s).

B_{0} = \frac{E_{0}}{c} = \frac{300}{3 \times 10^{8}} = 1 \times 10^{- 6} T

Step 2: Calculate maximum electric and magnetic forces○ Expand

The maximum electric force ( $F_{e}$ ) on an electron (charge $e$ ) is given by $F_{e} = e E_{0}$ . The electron moves in the y-direction ( $\vec{v} = v \hat{j}$ ) and the wave propagates in the x-direction. Since the electric field is in the y-direction, the magnetic field must be in the z-direction ( $\vec{B} = B_{0} \hat{k}$ ). The maximum magnetic force ( $F_{m}$ ) on the electron is given by the Lorentz force formula $F_{m} = | e (\vec{v} \times \vec{B}) |$ .

F_{e} = e E_{0}

\vec{v} \times \vec{B} = (v \hat{j}) \times (B_{0} \hat{k}) = v B_{0} (\hat{j} \times \hat{k}) = v B_{0} \hat{i}

F_{m} = e v B_{0}

💡 Teacher's Secret Hint

Remember the right-hand rule for vector cross products to determine the direction of the magnetic force.

Step 3: Determine the ratio of maximum electric to magnetic forces○ Expand

Now, we calculate the ratio of the maximum electric force to the maximum magnetic force. Substitute the expressions for $F_{e}$ and $F_{m}$ , and use the relationship $B_{0} = E_{0} / c$ .

\frac{F_{e}}{F_{m}} = \frac{e E_{0}}{e v B_{0}} = \frac{E_{0}}{v B_{0}}

\frac{F_{e}}{F_{m}} = \frac{E_{0}}{v (\frac{E_{0}}{c})} = \frac{c}{v}

Given the speed of the electron $v = 1.5 \times 10^{6}$ m/s and the speed of light $c = 3 \times 10^{8}$ m/s, the ratio is:

\frac{F_{e}}{F_{m}} = \frac{3 \times 10^{8} m/s}{1.5 \times 10^{6} m/s} = \frac{300 \times 10^{6}}{1.5 \times 10^{6}} = \frac{300}{1.5} = 200

💡 Teacher's Secret Hint

Notice that the charge of the electron and the electric field strength cancel out, simplifying the calculation significantly.

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Physics Question 43 – JEE-MAIN 2026

Clue 1 of 3: Understand the forces

🧠 Full Solution Path