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Electromagnetic Waves question

2021 · 31 Aug · Shift 2 · Q54
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  5. /2021 · 31 Aug · Shift 2 · Q54

Electromagnetic Waves question

2021 · 31 Aug · Shift 2 · Q54

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
The magnetic field vector of an electromagnetic wave is given by B=B0i^+j^2cos⁡(kz−ωt)B = {B_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos (kz - \omega t)B=B0​2​i+j​​cos(kz−ωt); where i^,j^\widehat i,\widehat ji,j​ represents unit vector along x and y-axis respectively. At t = 0s, two electric charges q1 of 4 π\piπ coulomb and q2 of 2 π\piπ coulomb located at (0,0,πk)\left( {0,0,{\pi \over k}} \right)(0,0,kπ​) and (0,0,3πk)\left( {0,0,{{3\pi } \over k}} \right)(0,0,k3π​), respectively, have the same velocity of 0.5 c i^\widehat ii, (where c is the velocity of light). The ratio of the force acting on charge q1 to q2 is :-
  1. A
    22:12\sqrt 2 :122​:1
  2. B
    1:21:\sqrt 21:2​
  3. C
    2 : 1
  4. D
    2:1\sqrt 2 :12​:1
View written solutionFree

Correct answer: C

  1. Given magnetic field

The magnetic field of the EM wave is

B⃗=B0i^+j^2cos⁡(kz−ωt)\vec B = B_0\frac{\hat i+\hat j}{\sqrt 2}\cos(kz-\omega t)B=B0​2​i^+j^​​cos(kz−ωt)

So its components are

Bx=By=B02cos⁡(kz−ωt)B_x=B_y=\frac{B_0}{\sqrt 2}\cos(kz-\omega t)Bx​=By​=2​B0​​cos(kz−ωt)

At t=0t=0t=0:

B⃗(z,0)=B0i^+j^2cos⁡(kz)\vec B(z,0)=B_0\frac{\hat i+\hat j}{\sqrt 2}\cos(kz)B(z,0)=B0​2​i^+j^​​cos(kz)
  1. Find magnetic field at the two charge positions

For q1q_1q1​ at z=π/kz=\pi/kz=π/k

cos⁡(kz)=cos⁡(k⋅πk)=cos⁡π=−1\cos(kz)=\cos\left(k\cdot \frac{\pi}{k}\right)=\cos \pi=-1cos(kz)=cos(k⋅kπ​)=cosπ=−1

Hence,

B⃗1=−B0i^+j^2\vec B_1=-B_0\frac{\hat i+\hat j}{\sqrt 2}B1​=−B0​2​i^+j^​​

For q2q_2q2​ at z=3πkz=\frac{3\pi}{k}z=k3π​

cos⁡(kz)=cos⁡(k⋅3πk)=cos⁡3π=−1\cos(kz)=\cos\left(k\cdot \frac{3\pi}{k}\right)=\cos 3\pi=-1cos(kz)=cos(k⋅k3π​)=cos3π=−1

Hence,

B⃗2=−B0i^+j^2\vec B_2=-B_0\frac{\hat i+\hat j}{\sqrt 2}B2​=−B0​2​i^+j^​​

So both particles experience the same magnetic field.

  1. Find electric field direction of the EM wave

For an electromagnetic wave,

E⃗⊥B⃗,E⃗×B⃗ is along direction of propagation\vec E \perp \vec B, \qquad \vec E\times \vec B \text{ is along direction of propagation}E⊥B,E×B is along direction of propagation

Since the phase is (kz−ωt)(kz-\omega t)(kz−ωt), the wave propagates along +k^+\hat k+k^ (positive zzz-direction).

Let

E⃗=Exi^+Eyj^\vec E=E_x\hat i+E_y\hat jE=Ex​i^+Ey​j^​

with E⃗⋅B⃗=0\vec E\cdot \vec B=0E⋅B=0.

Because

B⃗∝i^+j^,\vec B \propto \hat i+\hat j,B∝i^+j^​,

E⃗\vec EE must be along

i^−j^\hat i-\hat ji^−j^​

(or opposite sign).

Take

E⃗=cB0i^−j^2cos⁡(kz−ωt)\vec E = cB_0\frac{\hat i-\hat j}{\sqrt 2}\cos(kz-\omega t)E=cB0​2​i^−j^​​cos(kz−ωt)

which satisfies

E⃗×B⃗∥k^.\vec E\times \vec B \parallel \hat k.E×B∥k^.

At both charge positions, since cos⁡(kz)=−1\cos(kz)=-1cos(kz)=−1,

E⃗1=E⃗2=−cB0i^−j^2\vec E_1=\vec E_2=-cB_0\frac{\hat i-\hat j}{\sqrt 2}E1​=E2​=−cB0​2​i^−j^​​

Thus both charges also experience the same electric field.

  1. Lorentz force on each charge

Velocity of both charges:

v⃗=0.5c i^\vec v=0.5c\,\hat iv=0.5ci^

Lorentz force:

F⃗=q(E⃗+v⃗×B⃗)\vec F=q(\vec E+\vec v\times \vec B)F=q(E+v×B)

Since both particles have the same E⃗\vec EE, same B⃗\vec BB, and same v⃗\vec vv, the vector quantity

E⃗+v⃗×B⃗\vec E+\vec v\times \vec BE+v×B

is identical for both.

Therefore,

F1F2=q1q2\frac{F_1}{F_2}=\frac{q_1}{q_2}F2​F1​​=q2​q1​​

provided the common factor is nonzero. Here it is indeed nonzero.

Now,

q1=4π,q2=2πq_1=4\pi, \qquad q_2=2\piq1​=4π,q2​=2π

so

F1F2=4π2π=2\frac{F_1}{F_2}=\frac{4\pi}{2\pi}=2F2​F1​​=2π4π​=2

Hence the ratio is

2:12:12:1
  1. Check options
  • A: 22:12\sqrt2:122​:1 ❌
  • B: 1:21:\sqrt21:2​ ❌
  • C: 2:12:12:1 ✅
  • D: 2:1\sqrt2:12​:1 ❌

So the correct option is C.

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