Moving Charges and Magnetism

Moving Charges and Magnetism

Moderate Questions
Difficult Questions

Electricity and Magnetism

Current Electricity
Moving Charges and Magnetism
Magnetism and Matter
Electromagnetic Induction
Alternating Current
Electromagnetic Waves
  1. In a certain region of space electric field →E and magnetic field →B are perpendicular to each other and an electron enters in region perpendicular to the direction of →B and →E both and moves un-deflected, then velocity of electron is​








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    Electron moves undeflected if force exerted due to electric field is equal to force due to magnetic field. ​

    q|→v||→B| = q|→E| ⇒ |→v| =
    |→E|
    |→B|

    Correct Option: A

    Electron moves undeflected if force exerted due to electric field is equal to force due to magnetic field. ​

    q|→v||→B| = q|→E| ⇒ |→v| =
    |→E|
    |→B|

  1. A charged particle of charge q and mass m enters perpendicularly in a magnetic field . Kinetic energy of the particle is E; then frequency of rotation is​​








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    For circular path in magnetic field, mrω² = qvB

    ⇒ ω² =
    qvB
    		As v = rω
    mr

    ∴ ω² =
    q(rω)B
    		 ⇒ ω =
    qB
    mrm

    ​ ∴ If ν is frequency of roatation, then ​
    v =
    ω
    		 ⇒ v =
    qB
    2π2πm

    Correct Option: B

    For circular path in magnetic field, mrω² = qvB

    ⇒ ω² =
    qvB
    		As v = rω
    mr

    ∴ ω² =
    q(rω)B
    		 ⇒ ω =
    qB
    mrm

    ​ ∴ If ν is frequency of roatation, then ​
    v =
    ω
    		 ⇒ v =
    qB
    2π2πm

  1. A proton moving with a velocity 3 × 105 m/s enters a magnetic field of 0.3 tesla at an angle of 30º with the field. The radius of curvature of its path will be (e/m for proton = 108 C/kg)​








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    r =
    mv sinθ
    		 =
    3 × 105 sin 30°
    Be0.3 × 108

    3 × 105 × (1/2)
    		 = 0.5 × 10-2 m = 0.5 cm.
    3 × 107

    Correct Option: B

    r =
    mv sinθ
    		 =
    3 × 105 sin 30°
    Be0.3 × 108

    3 × 105 × (1/2)
    		 = 0.5 × 10-2 m = 0.5 cm.
    3 × 107

  1. When a proton is accelerated through 1 V, then its kinetic energy will be








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    Potential difference (V) = 1V, ​
    K.E. acquired = qV 
    ​= 1.6 × 10-19 × 1 ​
    = 1.6 × 10-19 joules  = 1 eV

    Correct Option: C

    Potential difference (V) = 1V, ​
    K.E. acquired = qV 
    ​= 1.6 × 10-19 × 1 ​
    = 1.6 × 10-19 joules  = 1 eV

  1. A positively charged particle moving due east enters a region of uniform magnetic field directed vertically upwards. The particle will​​








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    ​In a perpendicular magnetic field, the path of a charged particle is a circle, and the magnetic field does not cause any change in energy.

    Correct Option: B

    ​In a perpendicular magnetic field, the path of a charged particle is a circle, and the magnetic field does not cause any change in energy.