MHT CET · Physics · Electromagnetic Induction
The magnetic flux near the axis and inside the air core solenoid of length \(60 \mathrm{~cm}\) carrying current ' \(I\) ' is \(1.57 \times 10^{-6} \mathrm{~Wb}\). Its magnetic moment will be \(\left[\mu_0=4 \pi \times 10^{-7}\right.\), SI unit and cross sectional area is very small as compared to length of solenoid.]
- A \(1 \mathrm{Am}^2\)
- B \(0.25 \mathrm{Am}^2\)
- C \(0.5\mathrm{Am}^2\)
- D \(0.75 \mathrm{Am}^2\)
Answer & Solution
Correct Answer
(D) \(0.75 \mathrm{Am}^2\)
Step-by-step Solution
Detailed explanation
Magnetic field inside the solenoid is given by
\(
\begin{aligned}
& \mathrm{B}=\frac{\mu_0 \mathrm{NI}}{\mathrm{L}} \\
& \therefore \frac{\phi}{\mathrm{A}}=\frac{\mu_0 \mathrm{NI}}{\mathrm{L}} \\
& \therefore \text { Magnetic moment, NIA }=\frac{\phi \mathrm{L}}{\mu_0} \\
& =\frac{1.5 \times 10^{-6} \times 0.6}{4 \times 3.14 \times 10^{-7}} \\
& =0.75 \mathrm{Am}^2
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{B}=\frac{\mu_0 \mathrm{NI}}{\mathrm{L}} \\
& \therefore \frac{\phi}{\mathrm{A}}=\frac{\mu_0 \mathrm{NI}}{\mathrm{L}} \\
& \therefore \text { Magnetic moment, NIA }=\frac{\phi \mathrm{L}}{\mu_0} \\
& =\frac{1.5 \times 10^{-6} \times 0.6}{4 \times 3.14 \times 10^{-7}} \\
& =0.75 \mathrm{Am}^2
\end{aligned}
\)
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