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Moving Charges and Magnetism — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsMoving Charges and Magnetism5 Marks
Question
  1. Derive an expression for the force between two long parallel current carrying conductors.
  2. Use this expression to define S.I. unit of current.
  3. A long straight wire AB carries a current I. A proton P travels with a speed v, parallel to the wire, at a distance d from it in a direction opposite to the current as shown in the figure. What is the force experienced by the proton and what is its direction?

✓

Answer

  1.  

Two long parallel conductors ‘a’ and ‘b’ are separated by a distance d and carry (parallel) currents Ia and Ib, respectively. The conductor ‘a’ produces, the same magnetic field Ba at all points along the conductor ‘b’.

$\text{B}_{a} = \frac{\mu_{0}\text{I}_{a}}{2\pi\text{d}}$

Fba, is the force on a segment L of ‘b’ due to ‘a’. The magnitude of this force is given by

Fba = Ib LBa

$=\frac{\mu_{0}\text{I}_{a}\text{I}_{b}}{2\pi\text{d}}\text{L}$

  1. The ampere is the value of that steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section, and placed one metre apart in vacuum, would produce on each of these conductors a force equal to 2 × 10–7 newton per metre of length.
  2. Magnetic field due to the straight wire AB at a perpendicular distance d from it.

$\text{B} = \frac{\mu_{0}\text{I}}{2\pi\text{d}}$

Therefore, force on proton moving with velocity ‘v’ perpendicular to B, is

$\text{f} = \text{qvB} = \frac{\mu_{0}\text{Iqv}}{2\pi\text{d}}$

Direction: Towards right.

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