Question

A 20 cm long conducting rod is set into pure translation with a uniform velocity of 10 cm s⁻¹ perpendicular to its length. A uniform magnetic field of magnitude 0.10 T exists in a direction perpendicular to the plane of motion. (a) Find the average magnetic force on the free electrons of the rod. (b) For what electric field inside the rod, the electric force on a free elctron will balance the magnetic force? How is this electric field created? (c) Find the motional emf between the ends of the rod.

Answer 1

Given:-

Length of the rod, l = 20 cm = 0.2 m

Velocity of the rod, v = 10 cm/s = 0.1 m/s

Magnetic field, B = 0.10 T

(a) The force on a charged particle moving with velocity v in a magnetic field is given by

\[\overrightarrow{F} = q\left( \overrightarrow{v} \times \overrightarrow{B} \right)\]

F = qvB sin θ

Here,

θ = 90^o

Now,

F = (1.6 × 10⁻¹⁹) × (1 × 10⁻¹) × (1 × 10⁻¹)

= 1.6 × 10⁻²¹ N

(b) The electrostatic force on the charged particle is qE.

Here,

qE = qvB

⇒ E = (1 × 10⁻¹ ) × (1 × 10⁻¹)

= 1 × 10⁻² V/m

It is created because of the induced emf.

(c) Motional emf between the ends of the rod, e = Bvl

⇒ e = 0.1 × 0.1 × 0.2

= 2 × 10⁻³ V