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: Understanding the Question:
The question gives the total magnetic flux ($\Phi$) near the axis inside a long air-core solenoid. We are provided with the solenoid's length ($L = 60\text{ cm} = 0.6\text{ m}$) and the permeability of free space ($\mu_0$). We need to determine the total magnetic dipole moment ($M$) of the solenoid.

Step 2: Key Formula or Approach:
1.

Magnetic Field inside a Solenoid ($B$): The uniform field near the axis inside a long solenoid is: $$B = \frac{\mu_0 N I}{L}$$ 2.

Magnetic Flux ($\Phi$): Flux is the product of the magnetic field and the cross-sectional area $A$: $$\Phi = B \cdot A = \frac{\mu_0 N I A}{L}$$ 3.

Magnetic Moment ($M$): The total magnetic dipole moment of a coil winding is defined as: $$M = N I A$$ Substituting $M$ into the flux equation gives $\Phi = \frac{\mu_0 M}{L}$, which can be rearranged to isolate $M$: $$M = \frac{\Phi \cdot L}{\mu_0}$$

Step 3: Detailed Explanation:
Identify and organize the given parameters in standard SI units: Length of the solenoid, $L = 60\text{ cm} = 0.6\text{ m}$ Core magnetic flux, $\Phi = 1.57 \times 10^{-6}\text{ Wb}$ Permeability constant, $\mu_0 = 4\pi \times 10^{-7} \approx 4 \times 3.14 \times 10^{-7} = 12.56 \times 10^{-7}\text{ T}\cdot\text{m/A}$ Notice that the value $1.57$ given for the flux is exactly half of $\pi$ ($3.14 / 2 = 1.57$). Let's rewrite $\Phi$ using $\pi$ to make terms cancel cleanly: $$\Phi = \frac{\pi}{2} \times 10^{-6}\text{ Wb}$$ Now, substitute these parameters into our rearranged magnetic moment formula: $$M = \frac{\left( \frac{\pi}{2} \times 10^{-6} \right) \cdot 0.6}{4\pi \times 10^{-7}}$$ Cancel out the transcendental constant $\pi$ from both the numerator and denominator: $$M = \frac{0.6 \times 10^{-6}}{2 \cdot 4 \times 10^{-7}} = \frac{0.6 \times 10^{-6}}{8 \times 10^{-7}}$$ Simplify the base-10 exponents ($\frac{10^{-6}}{10^{-7}} = 10^1 = 10$): $$M = \frac{0.6 \times 10}{8} = \frac{6}{8}$$ Reduce the fraction to its decimal form: $$M = \frac{3}{4} = 0.75\text{ Am}^2$$

Step 4: Final Answer:
The total magnetic moment of the solenoid is $0.75\text{ Am}^2$, which corresponds to option (D).