Question

Q.13
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

Answer:

you can take help from this solution whereever the number is change you can put your questions digits

Answer 2

Answer:

Magnitude of magnetic field = 3.771 × 10⁻³ T

Explanation:

Given:

• Length of the solenoid = 25 cm = 0.25 m
• Radius = 1 cm = 0.01 m
• Current = 3 A

To Find:

• Magnitude of magnetic field inside the solenoid

Solution:

Magnetic field at the centre of a solenoid is given by the formula,

$\tt B=\mu_0\:nI$

where B is the magnetic field, I is the current passing through the solenoid.

Here n is given by,

$\tt n=\dfrac{Total\:number\:of\:turns}{Length\:of\:the\:solenoid}$

Substitute the data,

$\tt n=\dfrac{250}{0.25}$

$\tt n=1000$

Now substituting in the above formula we get,

$\tt B=12.57\times 10^{-7}\times 1000\times 3$

$\tt \implies 3.771\times 10^{-3}\: T$

Hence the magnetic field inside the solenoid is 3.771 × 10⁻³ T.