CBSE Class 12 · 2012 · All India · Magnetic Effects of CurrentLong Answer

(a) Write the expression for the force

F

acting on a charged particle of charge

q

moving with velocity

v

in the presence of both an electric field

E

and a magnetic field

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

B

. Prove that the torque

τ

acting on the loop is

τ = M \times B

, where

M

is the magnetic moment of the loop.

Solution

(a) The Lorentz force is

F = q E + q (v \times B)

. The particle goes undeflected when the electric and magnetic forces cancel:

q E = q v B

, i.e.

v = \frac{E}{B}

(with

E

B

v

mutually perpendicular) — a velocity selector. (b) The forces on the two sides of length

ℓ

are

F = B I ℓ

, equal and opposite, forming a couple with arm

b sin θ

τ = B I ℓ \cdot b sin θ = B I A sin θ = M B sin θ, M = I ℓ b .

In vector form

τ = M \times B

CBSE Class 12 Board · 2012 · All India · Subjective