(a) Write the expression for the force, `vecF`, acting on a charged particle of charge 'q', moving with a velocity `vecV` in the presence of both electric field `vecE` and magnetic field `vecB`. Obtain the condition under which the particle moves undeflected through the fields.
(b) A rectangular loop of size `lxxb` carrying a steady current I is placed in a uniform magnetic field `vecB`. Prove that the torque `vectau` acting on the loop is given by
`vectau=vecmxxvecB," where "vecm` is the magnetic moment of the loop.

Explanation

Electric force on particle `vecF_(e)=qvecE`
Magnetic force on particle `vecF_(m)=q(vecvxxvecB)`
Total force `vecF=vecF_(e)+vecF_(m)`
`vecF=q(vecE+vecvxxvecB)`
If a charge particle enter's perpendicular to both the electric and magnetic fields then it may happen that the electric and magnetic forces cancel each and so parallel will pass undeflected.
In such a case, `vecF=0`
`rArr q(vecE+vecvxxvecB)=0 rArrvecE=-(vecvxxvecB)`
`rArr vecE=vecBxxvecv`
`rArr vecEBv sin theta=Bv`(when `theta=90^(@)`)
`rArr v=E/B`(when `v E ` and `B` are mutually perpendicular)
(b)Torque on a current carrying loop:Consider a rectangular loop `PQRS` of length `l` breadth `b` suspended in a uniform magnetic field `vecB`.The length of loop `=PQ=RS=b`. let at any instant the normal to the plane of loop make an angle `theta` with the direction of magnetic field `vecB` and `I` be the current in the loop.We know that a force acts on a current carrying wire placed in a magnetic field.Therefore, each side of the loop will experience a force.The net force and torque acting on the loop will be determined by the force acting on all sides of the loop.Suppose that the forces on side `PQ,QR,RS` and `SP` are `vecF_(1),vecF_(2)vecF_(3)` and `vecF_(4)` respectively. The sides `QR` and `SP` make angle `(90^(@)-0)` with the direction of magnetic field.Therefore each of the forces `vecF_(2)` and `vecF_(4)` acting on these sides has same magnitude `F=Bb sin (90^(@)-0) B/b cos theta`.According to Fleming,s left hand rule the forces `vecF_(2)` and `vecF_(4)` are equal and opposite but their line of action is same. Therefore these forces cancel each other `i.e`. the resultant of `vecF_(2)` and `vecF_(4)` is zero.
This sides `PQ` and `RS` of current loop are perpendicular to the magnetic field therefore the magnitude of each of forces `vecF_(1)` and `vecF_(3)` is `F=I//B sin 90^(@)=I//B`
According to Fleming's left hand rule the forces `vecF_(1)` and `vecF_(3)` acting on sides `PQ` and `RS` are equal and opposite .but their lines of action different therefore the resultant force of `vecF_(1)` and `vecF_(3)` is zero.but they form a couple called the deflecting couple When the normal to plane of loop makes an angle `theta` with the direction of magnetic `B` the perpendicular distance between `vecF_(1)` and `vecF_(3)` is `b sin theta`.
`therefore` Moment of couple or Torque.
`tau=("Magnitude of one force" F)xx "perpendicular distance"=(BI//).(b sin theta) =I(b) B sin theta`
But b=area of loop =A (say)
`therefore` Torque `tau=IABsin theta`
If the loop contains `N`-turns then `tau=NiABsin theta`
In vector form `vectau=NI vecAxxvecB`
The magnetic dipole moment of rectangular current loop `=M=NIA`
therefore `vectau=vecM xx vecB`
Direction of torque is perpendicular to direction of area of loop as well as the direction of magnetic field i.e. along `I=vecA xx vecB`.