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Two toroids 1 and 2 have total no. of turns 200 and 100 respectively with average radii $40 \mathrm{~cm}$ and $20 \mathrm{~cm}$ respectively. If they carry same current i, the ratio of the magnetic fields along the two loops is, (1) $1: 2$ (2) $1: 1$ (3) $4: 1$ $(4) 2: 1$
Two toroids 1 and 2 have total no. of turns 200 and 100 respectively with average radii $40 \mathrm{~cm}$ and $20 \mathrm{~cm}$ respectively. If they carry same current i, the ratio of the magnetic fields along the two loops is, (1) $1: 2$ (2) $1: 1$ (3) $4: 1$ $(4) 2: 1$
Physics | Physics
Two toroids 1 and 2 have total no. of turns 200 and 100 respectively with average radii $40 \mathrm{~cm}$ and $20 \mathrm{~cm}$ respectively. If they carry same current i, the ratio of the magnetic fields along the two loops is, (1) $1: 2$ (2) $1: 1$ (3) $4: 1$ $(4) 2: 1$

Answer (2)
Sol. Magnetic field inside a toroid in
$
\mathrm{B}=\frac{\mu_{0} \mathrm{~N} \cdot \mathbf{I}}{2 \pi \mathrm{R}}
$
Here, $\frac{\mathrm{B}_{1}}{\overline{\mathrm{B}_{2}}}=\frac{\mathrm{N}_{1} \mathrm{R}_{2}}{\mathrm{~N}_{2} \mathbf{R}_{1}}=\frac{200}{100} \frac{20}{40}=1$
So, $\frac{B_{1}}{B_{2}}=1$

i

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