Question

A solenoid of radius 2.0 mm contains 100 turns of wire uniformly distributed over a length of 5.0 cm. It is located in air and carries a current of 2.0 A (i) Using Ampere's Law, calculate the magnetic field strength B inside the solenoid 4 marks] (ii) Draw a diagram clearly showing the direction of current flow in the solenoid and the direction of the magnetic field.

Answer 1

The magnetic field strength inside the solenoid is
`5.026*10^(-3)\ T`.

Step-by-step explanation:

Given that,

Radius = 2.0 mm

Length = 5.0 cm

Current = 2.0 A

Number of turns = 100

(a). We need to calculate the magnetic field strength inside the solenoid

Using formula of the magnetic field strength

Using Ampere's Law

`B=(\mu_(0)NI)/(l)`

Where, N = Number of turns

I = current

l = length

Put the value into the formula

`B=(4\pi*10^(-7)*100*2.0)/(5.0*10^(-2))`

`B=0.005026=5.026*10^(-3)\ T`

(b). We draw the diagram

Hence, The magnetic field strength inside the solenoid is
`5.026*10^(-3)\ T`.