Two toroids with number of turns 400 and 200 | vedclass

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Question

(D) The magnetic field $B$ inside a toroid is given by the formula:$B = \mu_0 \left( \frac{N}{2 \pi R} \right) I$where $N$ is the number of turns,$R$

Vedclass · Accepted Answer

$4:1$

Vedclass · Answer

(D) The magnetic field $B$ inside a toroid is given by the formula:$B = \mu_0 \left( \frac{N}{2 \pi R} ight) I$where $N$ is the number of turns,$R$ is the average radius,and $I$ is the current.From the formula,we see that $B \propto \frac{N}{R}$.Given:$N_1 = 400, R_1 = 30 \ cm$$N_2 = 200, R_2 = 60 \ cm$Since the current $I$ is the same for both,the ratio of the magnetic fields is:$\frac{B_1}{B_2} = \left( \frac{N_1}{N_2} ight) 	imes \left( \frac{R_2}{R_1} ight)$$\frac{B_1}{B_2} = \left( \frac{400}{200} ight) 	imes \left( \frac{60}{30} ight)$$\frac{B_1}{B_2} = 2 	imes 2 = 4$Therefore,the ratio is $4:1$.

Two toroids with number of turns $400$ and $200$ have average radii $30 \ cm$ and $60 \ cm$ respectively. If they carry the same current,the ratio of magnetic fields in these two toroids is:

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