Using formula B=2(R2+X2)3/2μ0II , we get 54=2[(3)2+(4)2]3/2μ0i(3)2 At the centre of the coil, X=0 and B=2(3)μ0i Using equation (i) B=(3)2×354×53⇒B=250μT.
The magnetic field due to a current carrying circular loop of radius 3 ~cm at a point on the axis at a distance of 4 ~cm from the centre is 54 μ T.…
JEE Main 2004 — Physics Electromagnetism
The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54μT. What will be its value at the centre of the Loop?
Held on 30 Apr 2004 · Verified 6 Jul 2026.
Options
- A
250μT
- B
150μT
- C
125μT
- D
75μT
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