JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A long straight wire with a circular crosssection having radius \(R\), is carrying a steady current \(I\). The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance \(r\;( r < R )\) from its centre will be
- A \(B \propto r^{2}\)
- B \(B \propto r\)
- C \(B \propto \frac{1}{ r ^{2}}\)
- D \(B \propto \frac{1}{r}\)
Answer & Solution
Correct Answer
(B) \(B \propto r\)
Step-by-step Solution
Detailed explanation
Use Ampere's law \(\text { B. } 2 \pi r =\mu_{0} \cdot \frac{ I }{\pi R ^{2}} \cdot \pi r ^{2}\) Thus \(B \propto r\)
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