A Rowland ring of mean radius 15; cm has 3500 | vedclass

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(C) The mean radius of the Rowland ring is $r = 15\; cm = 0.15\; m$.The number of turns on the ferromagnetic core is $N = 3500$.The relative permeabil

Vedclass · Accepted Answer

$4.48$

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(C) The mean radius of the Rowland ring is $r = 15\; cm = 0.15\; m$.The number of turns on the ferromagnetic core is $N = 3500$.The relative permeability of the core material is $\mu_r = 800$.The magnetizing current is $I = 1.2\; A$.The magnetic field $B$ inside a toroid (Rowland ring) is given by the formula:$B = \frac{\mu_r \mu_0 N I}{2 \pi r}$Where $\mu_0$ is the permeability of free space,$\mu_0 = 4 \pi 	imes 10^{-7}\; T\cdot m/A$.Substituting the values:$B = \frac{800 	imes (4 \pi 	imes 10^{-7}) 	imes 3500 	imes 1.2}{2 \pi 	imes 0.15}$$B = \frac{800 	imes 2 	imes 10^{-7} 	imes 3500 	imes 1.2}{0.15}$$B = \frac{800 	imes 2 	imes 10^{-7} 	imes 4200}{0.15} = \frac{0.672}{0.15} = 4.48\; T$.Thus,the magnetic field in the core is $4.48\; T$.

$A$ 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 $T$) in the core for a magnetizing current of $1.2\; A$?

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