Question

A cylindrical iron core supports N tums. If a current
I produces a magnetic flux (o) across the core's cross
section, then the magnetic energy is :-

(1) Io
2) 1/2 Io
$(3) {i}^{2} o$
(4)
${i}^{2} o \ \times (1 \div 2)$
​

Answer 1

Given that,

Numbers of turns = N

Current = I₀

magnetic flux = Φ

We know that,

Magnetic field is defined as,

$B=\dfrac{\mu_{0}NI_{0}}{2\pi r}$

Where, N = number of turns

I = current

r = radius

We need to calculate the magnetic energy

Using formula of magnetic energy

$\mu_{B}=\dfrac{B^2}{2\mu_{0}}$

Put the value into the formula

$\mu_{B}=\dfrac{(\dfrac{\mu_{0}NI_{0}}{2\pi r})^2}{2\mu_{0}}$

$\mu_{B}=\dfrac{\mu_{0}I_{0}^2N^2}{8\pi^2r^2}$  

$\mu_{B}\propto I_{0}^2$

Hence, The magnetic energy is proportional to I²₀.

(3) is correct option.