MHT CET · Physics · Magnetic Effects of Current
A long solenoid carrying a current produces a magnetic field B along its axis. If the number of turns per \(\mathrm{cm}\) is doubled and the current is made, \(\left(\frac{1}{3}\right)^{\text {rd }}\) then the new value of the magnetic field will be
- A \(\frac{B}{3}\)
- B \(3 \mathrm{~B}\)
- C \(2 \mathrm{~B}\)
- D \(\frac{2 \mathrm{~B}}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{2 \mathrm{~B}}{3}\)
Step-by-step Solution
Detailed explanation
Magnetic field inside a solenoid is given by
\(
\mathrm{B}=\mu_0 \mathrm{nI}
\)
If \(\mathrm{n}\) is doubled and \(\mathrm{I}\) is made \(\left(\frac{1}{3}\right)^{\mathrm{rd}}\), the value of \(\mathrm{B}\) will become \(\frac{2}{3} \mathrm{~B}\)
\(
\mathrm{B}=\mu_0 \mathrm{nI}
\)
If \(\mathrm{n}\) is doubled and \(\mathrm{I}\) is made \(\left(\frac{1}{3}\right)^{\mathrm{rd}}\), the value of \(\mathrm{B}\) will become \(\frac{2}{3} \mathrm{~B}\)
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