Question

The magnetic flux near the axis and inside the air core solenoid of length 80 cm carrying current 'I' is $1.57 \times 10^{-6}$ Wb. Its magnetic moment will be [cross-sectional area of a solenoid is very small as compared to its length, $\mu_0 = 4\pi \times 10^{-7}$ SI unit] ($\pi = 3.14$)

• A. 0.25 $Am^2$
• B. 0.50 $Am^2$
• C. 1 $Am^2$
• D. 1.2 $Am^2$

Answer 1

Answer: (C)

1 $Am^2$

Answer 2

The magnetic flux through the solenoid is given by

$\Phi = BA$,

and the magnetic field inside an air-core solenoid is

$B = \mu_0 \frac{N}{L} I$.

Thus,

$\Phi = \mu_0 \frac{N}{L} I A$.

Notice that the solenoid’s magnetic moment is

$m = NIA$.

Express $m$ in terms of $\Phi$:

$m = NIA = \frac{L}{\mu_0}\Phi$.

Substitute the given values ($L = 80\,\text{cm}=0.80\,\text{m}$, $\mu_0 = 4\pi\times10^{-7}\,\text{H/m}$, and $\Phi = 1.57 \times 10^{-6}\,\text{Wb}$):

$m = \frac{0.80}{4\pi \times 10^{-7}} \times (1.57 \times 10^{-6})$.

Since $4\pi \times 10^{-7} \approx 1.256 \times 10^{-6}$, we have:

$m = \frac{0.80}{1.256 \times 10^{-6}} \times (1.57 \times 10^{-6}) = \frac{0.80 \times 1.57 \times 10^{-6}}{1.256 \times 10^{-6}}$.

The $10^{-6}$ cancels out:

$m = \frac{0.80 \times 1.57}{1.256} \approx \frac{1.256}{1.256} = 1.00\,\text{A·m}^2$.

Thus, the magnetic moment is $1\,\text{A·m}^2$.