Question

A solenoid 15cm long has 300 turns of wire. A current of 5A flows through it. What is the magnitude of magnetic field inside the solenoid?

Answer 1

Answer:

Explanation:

Given,

Length of solenoid, l = 15 cm

Number of turns, n = 300

Current flow, C = 5 A

To Find,

The Magnetic field inside the solenoid.

Formula to be used,

Magnetic field due to the straight solenoid, at the solenoid, is given by

B = μ₀nl

Solution,

Putting all the values, we get

B = μ₀nl

⇒ B = 4π × 10⁻⁷ × (300/0.15) × 5

⇒ B = 12.56 × 10⁻⁷ × 10⁴

⇒ B = 1.3 × 10⁻² T.

Hence, the magnitude of the magnetic field inside the solenoid is 1.3 × 10⁻² T.

Answer 2

Answer:

Given :-

A solenoid 15cm long has 300 turns of wire. A current of 5A flows through it

To Find :-

Magnitude

Solution :-

At first we know that

100 cm = 1 m

15 cm = 15/100 = 0.15 m

Number of turns = 300

Length of one turn = 300/0.15 = 2000 mm

Now

Magnitude = 4 × π × $10^{-7}$ × 2000 × 5

Magnitude = 4 × 22/7 × $10^{-7}$ × 10,000

Magnitude = 88/7 × $10^{-7}$× 10⁴

Magnitude = 12.56 × $10^{-7}$× 10⁴

Magnitude = 12.56 × $10^{-7 + 4}$

Magnitude = 1.3 × 10$^{-2}$