Question

What is the field inside a 2.00-m-long solenoid that has 2000 loopsand carries a 1600-A current?​

Answer 1

Answer:

magnetic needle placed in uniform

magnetic field has magnetic moment of

2 x 10-2 A m², and moment of inertia of

7.2 x 10-7 kg m². It performs 10 complete

oscillations in 6 s. What is the magnitude

of the magnetic field ?

Answer 2

Answer:

The magnetic field is 2.0096 Wb / $m^{2}$

Explanation:

We have been given a 2 m long solenoid that is having 2000 loops and it is carrying a current of 1600 A.

We have to find the field inside the solenoid by the following formula:

B = μ₀n$I$

Where we have μ₀ to be the permeability of the free space,

n is the no. of turns per unit length of the solenoid and $I$ is the current in the solenoid.

Here we have,

μ₀ = 4π x $10^{-7} N/ A^{2}$

n = $\frac{N}{l}$

Here N is the no. of turns and l is the length of the solenoid.

∴ n = $\frac{2000}{2}$

⇒ n = 1000 turns per unit length.

$I$ = 1600 A

∴ B = 4π × $10^{-7} N/ A^{2}$ × 1000 × 1600

⇒ B = 2.0096 Wb / $m^{2}$

Therefore the magnetic field is 2.0096 Wb / $m^{2}$