The magnetic field within the turns of toroid is B=2πrμ0NI where, N= number of turns, I= current in loops and r= radius of each turn Here, N1=200, N2=100,r1=40 cm,r2=20 cm and current I is same, then B2B1=2πr1μ0 N1I×μ0 N2I2πr2 Substituting the given values in the above relation, we get B2B1∴B1:B2=( N2N1)(r1r2)=(100200)(4020)=2×21=1=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 same current…
NEET UG 2019 — Physics Electromagnetism
2019mcqeasy
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 same current i, the ratio of the magnetic fields along the two loops is
Official previous-year question
Held on 30 Apr 2019 · Verified 9 Jul 2026.
Options
- A
1:1
- B
4:1
- C
2:1
- D
1:2
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