1. Two straight long parallel conductors carry currents I1 and I2… - Vidyadip

Question

Two straight long parallel conductors carry currents I₁ and I₂ in the same direction. Deduce the expression for the force per unit length between them.

Depict the pattern of magnetic field lines, around them.

A rectangular current carrying loop EFGH is kept in a uniform magnetic field as shown in the figure.

What is the direction of the magnetic moment of the current loop?
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.
2. 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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