Question

Copper has 8 x 10^(28) electrons per cubic metre. A copper wire of length 1 m
and cross sectional area 8 x 10^(-6) m^2 carrying a current and lying at right angle
to a magnetic field of strength 5 x 10^(-3) T experiences a force of 8 x 10^(-2) N.
Calculate the drift velocity of free electrons in the wire.​

Answer 1

Explanation:

Given Copper has 8 x 10^(28) electrons per cubic metre. A copper wire of length 1 m and cross sectional area 8 x 10^(-6) m^2 carrying a current and lying at right angle  to a magnetic field of strength 5 x 10^(-3) T experiences a force of 8 x 10^(-2) N.

• Calculate the drift velocity of free electrons in the wire.
• We need to find the drift velocity.
• So n = 8 x 10^28 elec / m^3
• Length = 1 m
• Area = 8 x 10^-6 m^2
• B = 5 x 10^-3 T
• Force F = 8 x 10^-2 N
• Number of electrons present = 8 x 10^28 x 8 x 10^-6 x 1
•                                                  = 64 x 10^22
• Total charge present = 1.6 x 10^-19 x 64 x 10^22
•                                   = 102.4 x 10^3  
•                                   = 1.024 x 10^5 C
• So Force = q (V x B)
• Copper wire is lying at right angle to a magnetic field.
•              = q mod V x mod B sin 90
•             = q mod V mod B
• 8 x 10^-2 N = 1.024 x 10^5 x Vd (drift vel) x 5 x 10^-3
•      So Vd (drift velocity) = 8 x 10^-2 / 5 x 10^-3 x 1.024 x 10^5
•                 = 1.56 x 10^-4 m /s

Reference link will be

https://brainly.in/question/4121250

https://brainly.in/question/637700