MHT CET · Physics · Magnetic Effects of Current
A straight wire of diameter \(0.4 \mathrm{~mm}\) carrying a current of \(2 \mathrm{~A}\) is replaced by another wire of \(0.8 \mathrm{~mm}\) diameter carrying the same current. The magnetic field at distance (R) from both the wires is ' \(\mathrm{B}_1\) ' and ' \(\mathrm{B}_2\) ' respectively. The relation between \(\mathrm{B}_1\) and \(\mathrm{B}_2\) is
- A \(\mathrm{B}_1=\frac{\mathrm{B}_2}{2}\)
- B \(\mathrm{B}_1=\mathrm{B}_2\)
- C \(\mathrm{B}_1=2 \mathrm{~B}_2\)
- D \(\mathrm{B}_1=\frac{\mathrm{B}_2}{3}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{B}_1=\mathrm{B}_2\)
Step-by-step Solution
Detailed explanation
Since the same current flows through the second wire, the magnetic field at the same distance will be same.
\(
\therefore \mathrm{B}_1=\mathrm{B}_2
\)
\(
\therefore \mathrm{B}_1=\mathrm{B}_2
\)
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