A current of 1 ~A is flowing on the sides of an equilateral triangle of side 4.5 × 10^-2 ~m. The magnetic field at the centre of the triangle will be:
JEE Main 2018 — Physics Electromagnetism
2018mcqeasy
A current of 1A is flowing on the sides of an equilateral triangle of side 4.5×10−2m. The magnetic field at the centre of the triangle will be:
Official previous-year question
Held on 15 Apr 2018 · Verified 6 Jul 2026.
Options
A
4×10−5Wb/m2
B
Zero
C
2×10−5Wb/m2
D
8×10−5Wb/m2
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Solution
Here, side of the triangle, l=4.5×10−2m, current, I=1A magnetic field at the centre of the triangle ' O ' B= ? From figure, tan60∘=3=2d1⇒d=23l=(234.5×10−2)m
Magnetic field, B=4πdμ0i(cosθ1+cosθ2) Putting value of μ=4π×10−7 and θ1 and θ2 we will get net magenetic field =3×B=4×10−5Wb/m2
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