The magnetic field vector of an electromagnetic wave is given by $B=B_{0}=\frac{\hat{i}+\hat{j}}{\sqrt{2}} \cos (k z-\omega t)$; where $\hat{i}, \hat{j}$ represents unit vector along $x$ and $y$-axis respectively. At $t=0 \mathrm{~s}$, two electric charges q 1 of $4 \pi$ coulomb and q2 of $2 \pi$ coulomb located at $\left(0,0, \frac{\pi}{\mathrm{k}}\right)$ and $\left(0,0, \frac{3 \pi}{\mathrm{k}}\right)$, respectively, have the same velocity of $0.5 \mathrm{c} \hat{\mathrm{i}}$ (where c is the velocity of light). The ratio of the force acting on charge q1 to q2 is :-

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Magnetic Effects of Current

The magnetic field vector of an electromagnetic wave is given by

B = B_{0} = \frac{i ^ + j ^}{2} cos (k z - ω t)

; where

\hat{i}, \hat{j}

represents unit vector along

x

and

y

-axis respectively. At

t = 0 s

, two electric charges q 1 of

4 π

coulomb and q2 of

2 π

coulomb located at

(0, 0, \frac{π}{k})

and

(0, 0, \frac{3 π}{k})

, respectively, have the same velocity of

0.5 c \hat{i}

(where c is the velocity of light). The ratio of the force acting on charge q1 to q2 is :-

(A) $1 : 2$
(B) $2 : 1$
(C) $2 : 1$ ✓ Correct
(D) $22 : 1$

Answer: (C)

Solution

The wave field

B = B_{0} \frac{i ^ + j ^}{2} cos (k z - ω t)

has the same value at both charges at

t = 0

: at

z = π / k

cos π = - 1

, and at

z = 3 π / k

cos 3 π = - 1

. Both charges move with the same velocity

0.5 c \hat{i}

, and the wave's electric field shares the same spatial factor, so the total field acting on each charge is identical in magnitude and direction. The force on a charge is then proportional only to its charge, giving

\frac{F _{1}}{F _{2}} = \frac{q _{1}}{q _{2}} = \frac{4 π}{2 π} = 2.

The ratio of the forces is

2 : 1

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Magnetic Effects of Current

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