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. A milli voltmeter of 25 milli volt range is to be converted into an ammeter of 25 ampere range. The value (in ohm) of necessary shunt will be :








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    ​Galvanometer is converted into ammeter, by connected a shunt, in parallel with it. ​ ​ ​

    GS
    		 =
    VG
    		 =
    25 × 10-3
    G + SI25

    =
    GS
    		 = 0.001 Ω
    G + S

    ​Here S << G so
    ​S = 0.001 Ω

    Correct Option: A

    ​Galvanometer is converted into ammeter, by connected a shunt, in parallel with it. ​ ​ ​

    GS
    		 =
    VG
    		 =
    25 × 10-3
    G + SI25

    =
    GS
    		 = 0.001 Ω
    G + S

    ​Here S << G so
    ​S = 0.001 Ω

  1. A closely wound solenoid of 2000 turns and area of cross-section 1.5 × 10–4 m² carries a current of 2.0 A. It suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field 5 × 10–2 tesla making an angle of 30° with the axis of the solenoid. The torque on the solenoid will be:








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    Torque on the solenoid is given by τ ​= MB sin θ ​
    where θ is the angle between the magnetic field and the axis of solenoid. ​
    M = niA ​ 
    ∴ τ = niA B sin 30° ​ ​
    = 2000 × 2 × 1.5 × 10-4 × 5 × 10-2 × 1/2
    = 1.5 × 10-2 N - m

    Correct Option: D

    Torque on the solenoid is given by τ ​= MB sin θ ​
    where θ is the angle between the magnetic field and the axis of solenoid. ​
    M = niA ​ 
    ∴ τ = niA B sin 30° ​ ​
    = 2000 × 2 × 1.5 × 10-4 × 5 × 10-2 × 1/2
    = 1.5 × 10-2 N - m


  1. ​ ​A closed loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments PS, SR, and RQ are F1 , F2 and F3 respectively  and are in the plane of the paper and along the directions shown, the force on the segment QP is








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    According to the figure the magnitude of force on the segment QM is F3 – F1 and PM is F2. ​

    Therefore, the magnitude of the force on ​segment PQ is √(F3 - F1)² + F2²

    Correct Option: B

    According to the figure the magnitude of force on the segment QM is F3 – F1 and PM is F2. ​

    Therefore, the magnitude of the force on ​segment PQ is √(F3 - F1)² + F2²

  1. When a charged particle moving with velocity →v is subjected to a magnetic field of induction →B, the force on it is non-zero. This implies that​​








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    ​Force on a particle moving with velocity v in a magnetic field  B is F = q(→v × →B)
    ​If angle between →v & →B is either zero or 180º, then value of F will be zero as cross product of →v and →B will be zero. ​
    So option (b) is correct.

    Correct Option: B

    ​Force on a particle moving with velocity v in a magnetic field  B is F = q(→v × →B)
    ​If angle between →v & →B is either zero or 180º, then value of F will be zero as cross product of →v and →B will be zero. ​
    So option (b) is correct.

  1. A very long straight wire carries a current I. At the instant when a charge + Q at point P has velocity →c , as shown, the force on the charge is​​








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    The direction of →B is along (-̂k)
    ​∴ The magnetic  force ​ ​
    →F = Q(→v × →B) = Q(v̂i) × B(-̂k) = QvB̂j
    ⇒ →F is along OY.

    Correct Option: A

    The direction of →B is along (-̂k)
    ​∴ The magnetic  force ​ ​
    →F = Q(→v × →B) = Q(v̂i) × B(-̂k) = QvB̂j
    ⇒ →F is along OY.