Question

A long solenoid has n turns per unit length, has a radius R1, and carries a current R2. A circular coil with radius R2 and with N total turns is coaxial with the solenoid and equidistant from its ends. (a) Find the magnetic flux through the coil ifR2 > R1 (b) Find the magnetic flux through the coil if R2 < R1.

Answer 1

Answer:

a) Фm = μ₀*I*n*π*N*R₁²

b) Фm = μ₀*I*n*π*N*R₂²

Explanation:

Given

n = number of turns per unit length of the solenoid

R₁ = radius of the solenoid

I = current passing through the solenoid

R₂ = radius of the circular coil

N = number of total turns on the coil

a) Фm = ?   If  R₂ > R₁

b) Фm = ?   If  R₂ < R₁.

We can use the formula

Фm = N*B*S*Cos θ

where

N is the number of turns

B is the magnetic field

S is the area perpendicular to magnetic field B.

The magnetic field outside the solenoid can be approximated to zero. Therefore, the flux through the coil is the flux in the core of the solenoid.

We know that the magnetic field inside the solenoid is uniform. Thus, the flux through the circular coil is given by the same expression with R2 replacing R1 (from the area).

a)  Then, the flux through the large  circular loop outside the solenoid (R₂ > R₁) is obtained as follows:

Фm = N*B*S*Cos θ

where

B = μ₀*I*n

S = π*R₁²

θ = 0°

⇒ Фm = (N)*(μ₀*I*n)*(π*R₁²)*Cos 0°

⇒ Фm = μ₀*I*n*π*N*R₁²

b) The flux through the coil when  R₂ < R₁ is

Фm = (N)*(μ₀*I*n)*(π*R₂²)*Cos 0°

⇒ Фm = μ₀*I*n*π*N*R₂²

Answer 2

Answer:

a) ∅ $= u_0*n*I*N*pi*(R_1)^2$

b) ∅ $= u_0 * n*I*N * pi * (R_2)^2$

Explanation:

The magnetic field inside the solenoid is uniform and the flux in the coil is the same as the flux in the solenoid.

The magnetic field outside can be said to be zero which also means that the flux through the coil is the same as flux in the solenoid.

Thus, we can say

R2 = R1

Where:

N = total turns on coil

n = number of turns per unit length of solenoid

I = current

R1 = radius of solenoid

R2 = circular coil radius

a) The magnetic flux through the coil if R2 > R1 is giben as:

∅ $= N(u_0 *n*I)(pi*(R_1 )^2) Cos 0$

Solving further, we have:

∅ $= u_0*n*I*N*pi*(R_1)^2$

Here,

$u_0*n*I$ is the magnetic field

(Pi*R1²) is the area pependicular to magnetic field

N is the number of turns

b) The magnetic flux through the coil if R2 < R1

∅ $= N(u_0*n*I)(pi*(R_2)^2)Cos 0$

∅ $= u_0 * n*I*N * pi * (R_2)^2$

Here,

$u_0*n*I$ is the magnetic field

(Pi*R2²) is the area pependicular to magnetic field

N is the number of turns