Two toroids 1 and 2 have total number of turn | vedclass

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(A) The magnetic field $B$ inside a toroid is given by the formula $B = \frac{\mu_0 N i}{2 \pi r}$,where $N$ is the total number of turns,$i$ is the c

Vedclass · Accepted Answer

$1:1$

Vedclass · Answer

(A) The magnetic field $B$ inside a toroid is given by the formula $B = \frac{\mu_0 N i}{2 \pi r}$,where $N$ is the total number of turns,$i$ is the current,and $r$ is the average radius.Given for toroid $1$: $N_1 = 200$,$r_1 = 40 \; cm$.Given for toroid $2$: $N_2 = 100$,$r_2 = 20 \; cm$.Since the current $i$ is the same for both,the ratio of the magnetic fields is:$\frac{B_1}{B_2} = \frac{\frac{\mu_0 N_1 i}{2 \pi r_1}}{\frac{\mu_0 N_2 i}{2 \pi r_2}} = \frac{N_1}{N_2} 	imes \frac{r_2}{r_1}$Substituting the values:$\frac{B_1}{B_2} = \left( \frac{200}{100} ight) 	imes \left( \frac{20}{40} ight) = 2 	imes \frac{1}{2} = 1$Therefore,the ratio is $1:1$.

Two toroids $1$ and $2$ have total number of turns $200$ and $100$ respectively with average radii $40 \; cm$ and $20 \; cm$ respectively. If they carry the same current $i$,the ratio of the magnetic fields along the two loops is

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