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Moving Charges and Magnetism — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsMoving Charges and Magnetism5 Marks
Question
  1. Two straight long parallel conductors carry currents $I_1$ and $I_2$ in the same direction. Deduce the expression for the force per unit length between them.
Depict the pattern of magnetic field lines, around them.
  1. A rectangular current carrying loop EFGH is kept in a uniform magnetic field as shown in the figure.
  1. What is the direction of the magnetic moment of the current loop?
  2. When is the torque acting on the loop (A) maximum, (B) zero?
✓

Answer

Derivation of force per unit length:

Figure shows two long parallel conductors a and b separated by a distance d and carrying (parallel) currents $I_$a and $I_b,$ respectively. The conductor ‘a’ produces, the same magnetic field $B_a$ at all points along the conductor ‘b’. The right-hand rule tells us that the direction of this field is downwards (when the conductors are placed horizontally). From Ampere’s circuital law, its magnitude is given by
$\text{B}_{a} = \frac{\mu_{0}\text{I}_{a}}{2\pi\text{d}}$
The conductor ‘b’ carrying a current $I_b$ will experience a sideways force due to the field $B_a$. The direction of this force is towards the conductor ‘a’. We label this force as $F_{ba}$, the force on a segment L of ‘b’ due to ‘a’. The magnitude of this force is given by.
$\text{F}_{ba} = \text{I}_{b }\text{L B}_{a} = \frac{\mu_{0}\text{I}_{a}\text{I}_{b}}{2\pi\text{d}}\text{L}$
Let $f_{ba}$ represent the magnitude of the force $F_{ba}$ per unit length. Then, from the above equation
$\text{f}_{ba} = \frac{\mu_{0}\text{I}_{a}\text{I}_{b}}{2\pi\text{d}}$
Pattern of Magnetic Field lines:
    1. Direction of magnetic moment $\overrightarrow{\text{M}}$ of the current loop is perpendicular to the plane of the page and directed downwards.
      1. Torque acting on the loop is maximum when $\overrightarrow{\text{M}}$ is perpendicular dicular to $\overrightarrow{\text{B}}.$
      2. It is minimum when $\overrightarrow{\text{M}}$ is parallel to $\overrightarrow{\text{B}}.$

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