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 current loop in a magnetic field








  1. View Hint View Answer Discuss in Forum

    ​A current loop in a magnetic field is in equilibrium in two orientations one is stable and another unstable. ​
    τ ​= →M × →B = MB sin θ
    If θ = 0° ⇒ τ = 0 (stable) ​
    If θ = π ⇒ τ = 0 (unstable) ​

    Do not experience a torque in some orientations ​Hence option (c) is correct.

    Correct Option: C

    ​A current loop in a magnetic field is in equilibrium in two orientations one is stable and another unstable. ​
    τ ​= →M × →B = MB sin θ
    If θ = 0° ⇒ τ = 0 (stable) ​
    If θ = π ⇒ τ = 0 (unstable) ​

    Do not experience a torque in some orientations ​Hence option (c) is correct.

  1. A long straight wire carries a certain current and produces a magnetic field of
    2 × 10–4 =
    weber
    m²

    at a perpendicular distance of 5 cm from the wire. An electron situated at 5 cm from the wire moves with a velocity 107 m/s towards the wire along perpendicular to it. The force experienced by the electron will be
    ​(charge on electron =1.6 × 10–19 C)








  1. View Hint View Answer Discuss in Forum

    Given: ​Magnetic field B = 2 × 10–4 weber/m²
    ​Velocity of electron, v = 107  m/s
    ​Lorentz force F = qvB sin θ ​
    = 1.6 × 10–19 × 107 × 2 × 10–4 (∵ θ = 90°)
    ​= 3.2 × 10–16 N

    Correct Option: C

    Given: ​Magnetic field B = 2 × 10–4 weber/m²
    ​Velocity of electron, v = 107  m/s
    ​Lorentz force F = qvB sin θ ​
    = 1.6 × 10–19 × 107 × 2 × 10–4 (∵ θ = 90°)
    ​= 3.2 × 10–16 N

  1. A circular coil ABCD carrying a current i is placed in a uniform magnetic field. If the magnetic force on the segment AB is , →F the force on the remaining segment BCDA is








  1. View Hint View Answer Discuss in Forum

    ​Here, ​​→FAB + →FBCDA = →0
    = →FBCDA = - →FAB→F
    (∵ →FAB = →F)

    Correct Option: B

    ​Here, ​​→FAB + →FBCDA = →0
    = →FBCDA = - →FAB→F
    (∵ →FAB = →F)

  1. A current carrying loop in the form of a right angle isosceles triangle ABC is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is F, what is the force on the arm AC?​​








  1. View Hint View Answer Discuss in Forum

    Let a current i be flowing in the loop ABC in the direction shown in the figure. If the length of each of the sides AB and BC be x then ​
    |→F| = ixB

    ​where B is the magnitude of the magnetic force. ​The direction of →F  will be in the direction perpendicular to the plane of the paper and going into it.
    By Pythagorus theorem, ​AC = √x² + x² = √2x
    ​∴ Magnitude of force on AC
    ​= i √2 x B sin 45° ​ ​
    = i√2xB × 1/√2
    = ixb = ​|→F|
    The direction of the force on AC is perpendicular to the plane of the paper and going out of it. Hence, force on AC = - →F

    Correct Option: B

    Let a current i be flowing in the loop ABC in the direction shown in the figure. If the length of each of the sides AB and BC be x then ​
    |→F| = ixB

    ​where B is the magnitude of the magnetic force. ​The direction of →F  will be in the direction perpendicular to the plane of the paper and going into it.
    By Pythagorus theorem, ​AC = √x² + x² = √2x
    ​∴ Magnitude of force on AC
    ​= i √2 x B sin 45° ​ ​
    = i√2xB × 1/√2
    = ixb = ​|→F|
    The direction of the force on AC is perpendicular to the plane of the paper and going out of it. Hence, force on AC = - →F