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

Answer:

The magnetic field strength inside the solenoid is $5.026\times10^{-3}\ T$.

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

$B=\dfrac{\mu_{0}NI}{l}$

Where, N = Number of turns

I = current

l = length

Put the value into the formula

$B=\dfrac{4\pi\times10^{-7}\times100\times2.0}{5.0\times10^{-2}}$

$B=0.005026=5.026\times10^{-3}\ T$

(b). We draw the diagram

Hence, The magnetic field strength inside the solenoid is $5.026\times10^{-3}\ T$.