(a)

Write the expression for the force , acting on a charged particle of charge ‘q’, moving with a velocity in the presence of both electric field E and magnetic field B. Obtain the condition under which the particle moves un deflected through the fields.

(b)

A rectangular loop of size l × b carrying a steady current I is placed in a uniform magnetic field. Prove that the torque  acting on the loop is given by  is the magnetic moment of the loop.

a) Electric field on the particle is given by, 

Magnetic force on the particle is given by, 

Total force will be, 

When the charged particle moves perpendicular to both electric and magnetic field, F = 0

b) Consider a loop PQRS of length l, breadth b suspended in a uniform magnetic field.

Length of the loop, PQ= RS = l

Breadth of the loop, QR = SP = b
At any instance, plane of loop makes angle  with the direction of magnetic field B

Suppose that the forces on sides PQ, QR, RS and SP are   respectively.

The resultant of forces F₂ and F₄ is zero.

The side PQ and RS of current loop are perpendicular to the magnetic field, therefore the magnitude of each of forces F₁ and F₃ is,
F = I l B sin 90^o = I l B 
Moment of couple or torque is given by,
(Magnitude of one Force F) x perpendicular distance = (BIl). (b sin ) = I (lb) B sin 

But, lb is the area of the loop = A.

But, lb is the area of the loop = A.

In vector form, 

Magnetic dipole moment of rectangular current loop is given by, M = NIA

Direction of torque is perpendicular to the direction of magnetic field.