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Moving Charges and Magnetism — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsMoving Charges and Magnetism5 Marks
Question
  1. Derive the expression for the torque on a rectangular current carrying loop suspended in a uniform magnetic field.
  2. A proton and a deuteron having equal momenta enter in a region of uniform magnetic field at right angle to the direction of the field. Depict their trajectories in the field.
✓

Answer

  1.  

A rectangular loop $\text{ABCD}$ of dimensions $l$ and $b$, carrying a steady current is placed in uniform magnetic field as shown in fig; such that normal of the plane is at angle $q$ with the magnetic field lines.
The force $F_{BC}$ and $F_{AD}$ on arms $BC$ and $AD$ are equal, opposite and along the axis of the coil, so they cancel each other.
The forces $F_{AB}$ and $F_{CD}$ are also equal and opposite, but are not collinear, so they constitute a couple, and the magnitude of the torque can be given as
$\tau = \text{F}_{AB}.\frac{\text{b}}{2}\sin\theta + \text{F}_{CD}.\frac{\text{b}}{2}\sin\theta$
Since
$|\text{F}_{AB}| = |\text{F}_{CD}| =\text{BI}\ell$
$\tau = \text{BI}\ell\times\text{b}\times\sin\theta$
$ = \text{BI}(\ell\text{b})\sin\theta$
$= BI\ A \sin \theta$
$[\text{A} = \ell \text{b} = $ area of the rectangle
Since magnetic moment $m = I |A|$
$\tau = \text{mB} \sin \theta$
In vector from $\overrightarrow{\tau} = \overrightarrow{\text{m}}\times\overrightarrow{\text{B}}$
  1. If a charge particle enters right angle to the direction of magnetic field, it follows a circular trajectory, and radius can be given as
$\text{q}v\text{B} = \frac{\text{m}v^{2}}{\text{r}}$
$\Rightarrow\text{r} = \frac{\text{m}v}{\text{qB}} = \frac{\text{p}}{\text{qB}}$
Since momentum are equal, and they have equal charges.
So, $r_p : r_d = 1:1.$

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