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Electromagnetic Waves

JEE Main level, single correct, medium

9 of 165

Question 9 of 165

An electromagnetic wave travelling in $x$ -direction is described by field equation $E_y = 300\sin\omega\left(t - \dfrac{x}{c}\right)$ . 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 _______.

Show answer and solution

Answer: (A) $200$

The maximum electric force acting on the electron is

F_e = eE_0

.

The maximum magnetic force acting on the electron is

F_m = evB_0

, where

v

is the speed of the electron and

B_0

is the maximum magnetic field.

The ratio of the maximum electric force to the maximum magnetic force is given by

\dfrac{F_e}{F_m} = \dfrac{eE_0}{evB_0} = \dfrac{E_0}{vB_0}

.

For an electromagnetic wave, the ratio of the amplitudes of the electric and magnetic fields is

\dfrac{E_0}{B_0} = c

, where

c

is the speed of light in vacuum.

Substituting this into the ratio of forces yields

\dfrac{F_e}{F_m} = \dfrac{c}{v}

.

Given

c = 3 \times 10^8

m/s and

v = 1.5 \times 10^6

m/s, we get

\dfrac{F_e}{F_m} = \dfrac{3 \times 10^8}{1.5 \times 10^6} = 200

.

Answer:

200

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