Question

A solenoid of length 25 cm has inner radius of 1 cm and is made up of 250 turns of copper
wire, for a current of 3 A in it. What will be the magnitude of the magnetic field inside the
solenoid?​

Answer 1

Given info : A solenoid of length 25 cm has inner radius of 1 cm and is made up of 250 turns of copper  wire, for a current of 3 A in it.

To find : the magnitude of the magnetic field inside the solenoid is ...

solution :

• the mangetic field inside the solenoid doesn't depend on the diamter/radius of solenoid and position of observation point inside the solenoid.

• it is directly proportional to number of turns per unit length and current passing through solenoid.

the magnitude of magnetic field strength inside the solenoid is given by,$B=\mu_0nI$

where $\mu_0$ is permeability of medium. i.e., vacuum [ if specific medium doesn't mention ]

$\mu_0$ = 4π × 10⁻⁷ H/m

n is number of turns per unit length

n = no of turns/length of solenoid = 250/(25/100 m) = 1000 turns per m

I = current passing through it = 3A

now the magnetic field inside the solenoid, B = 4π × 10⁻⁷ H/m × 1000 /m × 3A = 3.77 × 10⁻³ T

therefore the magnitude of the magnetic field inside the solenoid is 3.77 × 10⁻³ T

Answer 2

Answer:

Given info : A solenoid of length 25 cm has inner radius of 1 cm and is made up of 250 turns of copper wire, for a current of 3 A in it.

To find : the magnitude of the magnetic field inside the solenoid is ...

solution :

the mangetic field inside the solenoid doesn't depend on the diamter/radius of solenoid and position of observation point inside the solenoid.

it is directly proportional to number of turns per unit length and current passing through solenoid.

the magnitude of magnetic field strength inside the solenoid is given by,B=\mu_0nIB=μ

0

nI

where \mu_0μ

0

is permeability of medium. i.e., vacuum [ if specific medium doesn't mention ]

\mu_0μ

0

= 4π × 10⁻⁷ H/m

n is number of turns per unit length

n = no of turns/length of solenoid = 250/(25/100 m) = 1000 turns per m

I = current passing through it = 3A

now the magnetic field inside the solenoid, B = 4π × 10⁻⁷ H/m × 1000 /m × 3A = 3.77 × 10⁻³ T

therefore the magnitude of the magnetic field inside the solenoid is 3.77 × 10⁻³ T