Question

A solenoid 81.0 cm long has a radius of 1.70 cm and a winding of 1300 turns; it carries a current of 3.60 A. Calculate the magnitude of the magnetic field inside the solenoid.

Answer 1

Answer:

The magnitude of the magnetic field inside the solenoid is $7.3\times10^{-3}\ T$.

Explanation:

Given that,

Length = 81.0 cm

Radius = 1.70 cm

Number of turns = 1300

Current = 3.60 A

We need to calculate the magnetic field

Using formula of magnetic field inside the solenoid

$B =\mu nI$

$B =\mu\dfrac{N}{l}I$

Where, $\dfrac{N}{l}$=Number of turns per unit length

I = current

B = magnetic field

Put the value into the formula

$B =4\pi\times10^{-7}\times\dfrac{1300}{81.0\times10^{-2}}\times3.60$

$B = 7.3\times10^{-3}\ T$

Hence, The magnitude of the magnetic field inside the solenoid is $7.3\times10^{-3}\ T$.