Question

A solenoid 0.425 m long has 950 turns of wire. What is the magnetic field in the center of the solenoid when it carries a current of 2.75 A? (JC 19.57)

Answer 1

Answer:

The magnetic field in the center of the solenoid is $7.8\times10^{-3}\ T$.

Explanation:

Given that,

Length of solenoid = 0.425 m

Number of turns N = 950

Current I = 2.75 A

The magnetic field in the center of the solenoid is the product of the current , number of turns per unit length and permeability.

In mathematical form,

$B = \mu_{0}nI$

Where, $n = \dfrac{N}{l}$

N = number of turns

L = length

I = current

Now, The magnetic field

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

Put the value into the formula

$B=\dfrac{4\pi\times10^{-7}\times950\times2.75}{0.425}$

$B=\dfrac{4\times3.14\times10^{-7}\times950\times2.75}{0.425}$

$B=7.8\times10^{-3}\ T$

Hence, The magnetic field in the center of the solenoid is $7.8\times10^{-3}\ T$.