A toroid has 500 turns per metre length. If i | vedclass

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(B) The magnetic field $B$ inside a toroid is given by $B = \mu_0 n I$, where $n$ is the number of turns per unit length and $I$ is the current.Given:

Vedclass · Accepted Answer

$0.628$

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(B) The magnetic field $B$ inside a toroid is given by $B = \mu_0 n I$, where $n$ is the number of turns per unit length and $I$ is the current.Given: $n = 500 	ext{ m}^{-1}$, $I = 2 	ext{ A}$, $\mu_0 = 4\pi 	imes 10^{-7} 	ext{ T m/A}$.The magnetic energy density $u_B$ is given by $u_B = \frac{B^2}{2\mu_0}$.Substituting $B = \mu_0 n I$ into the formula, we get $u_B = \frac{(\mu_0 n I)^2}{2\mu_0} = \frac{\mu_0 n^2 I^2}{2}$.Calculating the value: $u_B = \frac{(4\pi 	imes 10^{-7}) 	imes (500)^2 	imes (2)^2}{2}$.$u_B = \frac{(4\pi 	imes 10^{-7}) 	imes 250000 	imes 4}{2} = 2\pi 	imes 10^{-7} 	imes 10^6 = 2\pi 	imes 10^{-1} = 0.2 	imes 3.14 = 0.628 	ext{ J/m}^3$.

$A$ toroid has $500$ turns per metre length. If it carries a current of $2 \text{ A}$, the magnetic energy density inside the toroid is: (in $\text{ J/m}^3$)

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