Question

The magnetic flux near the axis and inside an air core solenoid of length $60\text{ cm}$ carrying current $I$ is $1.57 \times 10^{-6}\text{ Wb}$. Its magnetic moment will be [$\mu_0 = 4\pi \times 10^{-7}$ SI units, and cross-sectional area is very small as compared to length of solenoid]

• A. $1\text{ Am}^2$
• B. $0.25\text{ Am}^2$
• C. $0.5\text{ Am}^2$
• D. $0.75\text{ Am}^2$

Hint:

Always look for hidden multiples of $\pi$ in physics constants! Recognizing that $1.57 = \frac{\pi}{2}$ lets you cancel terms directly in your head and completely avoids tedious long division with decimals during the exam.

Answer 1

The Correct Option is D

Step 1: What is given.
A long air-core solenoid of length $L = 60$ cm $= 0.6$ m carries current $I$. The magnetic flux near its axis is $\Phi = 1.57 \times 10^{-6}$ Wb. We want its magnetic moment $M$.
Step 2: Field inside a solenoid.
The field near the axis is \[ B = \frac{\mu_0 N I}{L} \] where $N$ is the number of turns.
Step 3: Flux through the solenoid.
Flux is field times cross-section area $A$: \[ \Phi = B A = \frac{\mu_0 N I A}{L} \]
Step 4: Bring in the magnetic moment.
The magnetic moment of the coil is $M = N I A$. So \[ \Phi = \frac{\mu_0 M}{L} \;\Rightarrow\; M = \frac{\Phi L}{\mu_0} \]
Step 5: Write flux using pi.
Note $1.57$ is half of $3.14$, so $\Phi = \dfrac{\pi}{2}\times 10^{-6}$ Wb. With $\mu_0 = 4\pi \times 10^{-7}$: \[ M = \frac{(\pi/2)\times 10^{-6} \times 0.6}{4\pi \times 10^{-7}} \]
Step 6: Cancel pi and simplify.
The $\pi$ cancels: \[ M = \frac{0.6 \times 10^{-6}}{8 \times 10^{-7}} = \frac{0.6 \times 10}{8} = \frac{6}{8} = 0.75\ \text{Am}^2 \] This is option (4). \[ \boxed{M = 0.75\ \text{Am}^2} \]