Question

A coil of 600 turns is wound uniformly on a ring of non-magnetic material. The ring has a uniform cross-sectional area of 200mm^2 and a mean circumference of 500mm. If the current in the coil is 4A, determine the :
i. magnetic field strength [3]
ii. flux density [3]
iii. total magnetic flux in the ring

Answer 1

Pls Refer the attachment

I hope this helps ( ╹▽╹ )

Answer 2

Answer:

(a)Magnetic field strength -6.0288×10⁻³T

(b)Flux density-6.0288×10⁻³T

(c)Total magnetic flux in the ring-1.20576× 10⁻⁶Tm²

Explanation:

The values given in the question are,

Number of turns(N)=600

Area of the ring(A)=200mm²=200×10⁻⁶m²

The mean circumference(l)=500mm=500×10⁻³m

The current in the coil(I)=4A

(a)

The magnetic field strength is given as,

$B=\mu_onI$        (1)

Where,

B=magnetic field strength of the ring

μ₀=4π×10⁻⁷

n=number of turns per unit length of the ring

I=current flowing in the ring

First, let's find n i.e.

$n=\frac{600}{500\times 10^-^3}=1200$       (2)

By putting the value of μ₀,n, and I in equation (1) we get;

$B=4\pi\times 10^-^7\times 1200\times 4=192\pi \times 10^-^5=6.0288\times 10^-^3T$    ( ∴π=3.14)

(b)

The flux density will be the same as the magnetic field strength. i.e.$6.0288\times 10^-^3T$.

(c)

The total flux density is given as,

$\phi=B.A$                (3)

Where,

Ф=total flux density

B=magnetic field strength

A=area of the ring

By substituting the value of B and A in equation (3) we get;

$\phi=6.0288\times 10^-^3\times 200\times 10^-^6=12.0576\times 10^-^7=1.20576\times 10^-^6Tm^2$

Hence, the magnetic field strength is 6.0288×10⁻³T, the flux density is 6.0288×10⁻³T and the total magnetic flux in the ring is 1.20576× 10⁻⁶Tm².