Q6 / 15 · NTA Abhyas (JEE Main) 2020 · Physics · Electromagnetic Waves

Cathode rays of velocity 106 m s-1 describe an approximate circular path of the radius 1 m in an electric field 300 Vcm-1. If the velocity of cathode rays are doubled. The value of electric field so that the rays describe the same circular path, will be

  • A
    2400 V cm-1
  • B
    600 V cm-1
  • Correct answer
    C
    1200 V cm-1
  • D
    12000 V cm-1

Solution

Cathode rays are composed of electrons, when they move in electric field a force

F=eE    ...(i)

Acts on them, this provides the necessary centripetal force to the particles

F=mv2r   ...(ii)

From Eqs. (i) and (ii), we get

eE=mv2r

⇒    r=mv2eE=m1062e(300)     ...(iii)

When velocity is doubled same circular path is followed, hence radius is same

r=m2×1062eE   ...(iv)

Equating Eqs. (iii) and (iv), we get

m×1062300e=m×2×1062eE

⇒   E=300×4=1200 V cm-1

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

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