MHT CET · Physics · Magnetic Effects of Current
A long solenoid carrying a current produces magnetic field B along its axis. If the number of turns per cm are tripled and the current is made \(\left(\frac{1}{4}\right)^{\mathrm{th}}\) then the new value of magnetic field will be
- A \(\frac{B}{3}\)
- B \(\frac{\mathrm{B}}{4}\)
- C \(\frac{3 B}{4}\)
- D \(\frac{2 \mathrm{~B}}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{3 B}{4}\)
Step-by-step Solution
Detailed explanation
Magnetic field due to a solenoid is given by, \(B=\mu_0 \mathrm{NIL}\)
After reducing current and increasing number of turns,
\(\mathrm{B}^{\prime}=\mu_0(3 \mathrm{~N})\left(\frac{\mathrm{I}}{4}\right) \mathrm{L}=\frac{3}{4}\left(\mu_0 \mathrm{NIL}\right)=\frac{3}{4} \mathrm{~B}\)
After reducing current and increasing number of turns,
\(\mathrm{B}^{\prime}=\mu_0(3 \mathrm{~N})\left(\frac{\mathrm{I}}{4}\right) \mathrm{L}=\frac{3}{4}\left(\mu_0 \mathrm{NIL}\right)=\frac{3}{4} \mathrm{~B}\)
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