Question

(a) Write the expression for the force $ \overrightarrow F$ , acting on a charged particle of charge $q$ moving with a velocity $\overrightarrow v $ in the presence of electric field $\overrightarrow E $ and magnetic field $\overrightarrow B $ . Obtain the condition under which the particle moves undeflected through the fields.
(b) A rectangular loop of size $l \times b$ carrying a steady current $I$ is placed in a uniform magnetic field $\overrightarrow B $ .Prove that the torque $\overrightarrow \tau $ acting on the loop is given by $\overrightarrow \tau = \overrightarrow m \times \overrightarrow B.$

Answer 1

Hint: In the case of a positive charge, the force on a charged particle due to an electric field is oriented parallel to the electric field vector and, in the case of a negative charge, anti-parallel. It is not dependent on the particle's velocity. There was a mistake. To decide the position of the impact, the right hand rule may be used.
Complete step by step solution:
The force acting on charge $q$ can be defined as
\[F = q\overrightarrow E + q (v \times \overrightarrow B)\]
Where the magnetic field is $\overrightarrow B $ , velocity is $\overrightarrow v $ and electric field is $\overrightarrow E.$
Take into consideration an area of which magnetic, electric and charging particle velocity sections are perpendicular. The net force on the charge particle must be zero if load particles are not to be moved.
\[q\overrightarrow E = qv\overrightarrow B \]
Therefore,
\[v = \dfrac{{\overrightarrow E}} {{\overrightarrow B}}\]
The magnetic and electrical field direction is equal and opposite. The magnitude is thus cancelling each other out in order to give a net power of nil. This doesn't deflect the charging particle.
Let’s consider the rectangular loop $ABCD$ with length $l$ and breadth $b$ suspended in uniform field $\overrightarrow B $
$AB = CD = l$
$BC = AD = b$
Assuming that the forces acting on $AB, BC, CD$ and $AD$ are $\overrightarrow {{F_1}} $ , $\overrightarrow {{F_2}} $ , ${\overrightarrow F _3} $ and $\overrightarrow {{F_4}} $ respectively.
We know that,
$\overrightarrow {{F_2}} = \overrightarrow {{F_4}} = 0$ (Resultant forces are zero).
Also,
$\overrightarrow {{F_1}} = \overrightarrow {{F_3}} = IlB\sin \theta $
Since these forces are perpendicular to the magnetic field therefore,
$\theta = {90^0}$
Hence,
$\overrightarrow {{F_1}} = \overrightarrow {{F_3}} = IlB$
Torque can be defined as the product of magnitude of force and perpendicular distance,
$\tau = I (lb) B\sin \theta $
Considering the area of loop as $A$
$\tau = NIAB\sin \theta $
In vector form,
$\overrightarrow \tau = NI\overrightarrow E \times \overrightarrow B $
Magnetic dipole is given by
$\overrightarrow m = NIA$
Therefore, \[\overrightarrow \tau = \overrightarrow m \times \overrightarrow B \].
Note: A filled particle does not operate magnetic power. Therefore the momentum of a charged particle cannot be boosted by the magnetic field. The particle goes in a circular direction if v and B are perpendicular. There was a mistake. Sections of this pace are unchanged.