MHT CET · Physics · Magnetic Effects of Current
The magnetic flux near the axis and inside the air core solenoid of length 80 cm carrying current ' I ' is \(1.57 \times 10^{-6} \mathrm{~Wb}\). Its magnetic moment will be [cross-sectional area of a solenoid is very small as compared to its length,\(\mu_0=4 \pi \times 10^{-7} \text { SI unit] }(\pi=3.14)\)
- A \(0.25 \mathrm{Am}^2\)
- B \(0.50 \mathrm{Am}^2\)
- C \(1 \mathrm{Am}^2\)
- D \(1.2 \mathrm{Am}^2\)
Answer & Solution
Correct Answer
(C) \(1 \mathrm{Am}^2\)
Step-by-step Solution
Detailed explanation
The magnetic induction inside the solenoid,
\(B=\frac{\mu_0 \mathrm{NI}}{L}\)
Magnetic flux,
\(\phi=\mathrm{BA}=\frac{\mu_0 \mathrm{NIA}}{\mathrm{~L}}\)
\(\begin{aligned} \text { Magnetic moment } & =\mathrm{NIA}=\frac{\phi \mathrm{L}}{\mu_0} \\ & =\frac{\left(1.57 \times 10^{-6}\right) \times 0.8}{4 \pi \times 10^{-7}} \\ & =1 \mathrm{Am}^2\end{aligned}\)
\(B=\frac{\mu_0 \mathrm{NI}}{L}\)
Magnetic flux,
\(\phi=\mathrm{BA}=\frac{\mu_0 \mathrm{NIA}}{\mathrm{~L}}\)
\(\begin{aligned} \text { Magnetic moment } & =\mathrm{NIA}=\frac{\phi \mathrm{L}}{\mu_0} \\ & =\frac{\left(1.57 \times 10^{-6}\right) \times 0.8}{4 \pi \times 10^{-7}} \\ & =1 \mathrm{Am}^2\end{aligned}\)
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