Q73 / 90 · JEE Main 2021 (31 Aug Shift 2) · Physics

The magnetic field vector of an electromagnetic wave is given by B=B0i^+j^2coskz-ωt where i^, j^ represents unit vector along x and y-axis respectively. At t=0 s, two electric charges q1 of 4π coulomb and q2 of 2π coulomb located at 0, 0, πk and 0, 0, 3πk, respectively, have the same velocity of 0.5ci^, (where c is the velocity of light ). The ratio of the force acting on charge q1 to q2 is :

  • A
    1 : 2
  • B
    22 : 1
  • C
    2 : 1
  • Correct answer
    D
    2 : 1

Solution

At t=0
B at 0, 0, πk=B0i^+j^2cosπ
B at 0, 0, 3πk=B0i^+j^2cos3π
Force on charged particle q1
F1=q1V→1×B→1
=4π0.5ci^×-B0i^+j^2=4πB0c22-k^
Force on charged particle q2
F2=q2V→2×B→2
=2π0.5ci^×-B0i^+j^2=2πB0c22-k^⇒F1F2=2

Hence, the correct answer is option: 'D'.

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