COMEDK · Physics · 20. Magnetic Properties of Matter
A vertical circular coil of radius \(0.1 \mathrm{~m}\) and having \(10\) turns carries a steady current. When the plane of the coil is normal to the magnetic meridian, a neutral point is observed at the centre of the coil. If \(B_{H}=0.314 \times 10^{-4} \mathrm{~T}\), the current in the coil is
- A \(0.5 \mathrm{~A}\)
- B \(0.25 \mathrm{~A}\)
- C \(2 \mathrm{~A}\)
- D \(1 \mathrm{~A}\)
Answer & Solution
Correct Answer
(A) \(0.5 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
Given, \(R=0.1 \mathrm{~m}, N=10, B_{H}=0.314 \times 10^{-4} \mathrm{~T}\) For neutral point, the magnetic field at the centre of coil carrying current is equal to the horizontal component of earth's field.…
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