Question

A Rowland ring of mean radius $15\,cm$ has $3500$ turns of wire wound on a ferromagnetic core of relative permeability $800$. What is the magnetic field B in the core for a magnetising current of $1.2\,A$ ?
(A) $4.8$
(B) $6$
(C) $7$
(D) $8$

Answer 1

Hint Ferromagnet is a simple tool with which some metals like the iron configure itself into a permanent magnet, or they get attracted towards the magnet. Some of the ferromagnetic materials are iron nickel and cobalt. A ferromagnetic material carries permanent magnetic atomic dipoles that are impulsively aligned corresponding to one another even in non-attendance of an outside the field.

Useful formula
By using the formula for the magnetic field of the toroid;
$B = \mu nI$
Where, $B$ denotes the magnetic field, $\mu $ denotes the relative permeability, $n$ denotes the number turns per unit area, $I$ denotes the magnetising current.

Complete step-by-step answer:
The data given in the problem is;
Radius of the ring $r = 15\,cm$,
Number of turns on the ring $N = \,3500$,
Relative permeability of the core ${\mu _r} = 800$,
Magnetising current $I = 1.2\,A$.

By using the formula for the magnetic field of the toroid;
$B = \mu nI$
The number of turns per unit length is;
$n = \dfrac{N}{{2\pi r}}$
Where, $N$ denotes the number of turns on the ring.
Substitute the values of $N$ and $r$ in the above equation;
Change the radius of the ring from $cm$ to $m$ .that is;
$r = 0.15\,cm$ ;
$n = \dfrac{{3500}}{{2 \times \pi \times 0.15}}$
But
$\mu = {\mu _0}{\mu _r}$
Where, ${\mu _0}$ denotes the relative permeability of free space, ${\mu _r}$ denotes the relative permeability of the core.
We know that that the relative permeability of free space ${\mu _0} = 4\pi \times {10^{ - 7}}\,N{A^{ - 2}}$ ;
$B = \mu nI$
Substitute the value of number turns of the ring per unit area $n$ and the relative permeability $\mu $, magnetising current $I$ ;
$B = \dfrac{{{\mu _0}{\mu _r} \times N \times I}}{{2\pi r}}$
$B = \dfrac{{800 \times 4\pi \times {{10}^{ - 7}} \times 3500 \times 1.2}}{{2\pi \times 0.15}}$

Simplify the equation;
$B = \dfrac{{33,60,000 \times 12.5663 \times {{10}^{ - 7}}}}{{0.9424777}}$
$B = 4.48\,T$

Therefore, the magnetic field of the core is $B = 4.48\,T$ .

Hence, the option (A) $4.8$ is the correct answer.

Note: Field of magnet is a vector field quantity in which the two of direction and the magnitude are present. The S.I. unit is a Tesla that is $T$. The evaluation of the resistance of a particular material against the emergence of in the transmission line theory is known as the permeability. The permeability of free space is a physical constant which is frequently used in electromagnetism.