Question 265: Magnetism and Matter

Question

A straight magnetic strip has a magnetic moment of $$44 \mathrm{~Am}^2$$. If the strip is bent in a semicircular shape, its magnetic moment will be ________  $$\mathrm{Am}^2$$.
(given $$\pi=\frac{22}{7}$$)

Accepted Answer

Magnetic moment is defined as the product of the magnet's pole strength and the distance between the poles (also known as the magnetic length). When a magnetic strip is bent, its magnetic moment changes based on the new configuration.
Consider a straight magnetic strip with a magnetic moment of $$44 \, 	ext{Am}^2$$. If this strip is bent into a semicircular shape, we need to find the new effective magnetic moment.
The magnetic moment in a straight strip is given by:
$$ M_{	ext{straight}} = m \cdot l $$
where:
$$m$$ is the pole strength
$$l$$ is the magnetic length
Given $$M_{	ext{straight}} = 44 \, 	ext{Am}^2$$, let's now consider the strip bent into a semicircular shape.
When the strip is bent into a semicircle, the effective distance between the magnetic poles is the diameter of the semicircle. Let's denote the original length of the strip as $$L$$. In a straight line, this length $$L$$ is also the magnetic length. When bent into a semicircle, the length of the arc of the semicircle is still $$L$$.
The circumference of a full circle is given by:
$$ C = 2\pi R $$
Therefore, the length of the arc of a semicircle is:
$$ L = \pi R $$
Solving for $$R$$, we get:
$$ R = \frac{L}{\pi} $$
The diameter of the semicircle (which is the new effective magnetic length, $$l_{	ext{new}}$$) is twice the radius:
$$ l_{	ext{new}} = 2R = 2 \cdot \frac{L}{\pi} = \frac{2L}{\pi} $$
Now, the new magnetic moment $$M_{	ext{new}}$$ is:
$$ M_{	ext{new}} = m \cdot l_{	ext{new}} = m \cdot \frac{2L}{\pi} $$
We know from the original magnetic strip:
$$ M_{	ext{straight}} = m \cdot L = 44 \, 	ext{Am}^2 $$
Rewriting $$m$$ in terms of the known magnetic moment of the straight strip:
$$ m = \frac{44}{L} $$
Substituting $$m$$ into the new magnetic moment equation:
$$ M_{	ext{new}} = \frac{44}{L} \cdot \frac{2L}{\pi} $$
Canceling out $$L$$ from the numerator and the denominator:
$$ M_{	ext{new}} = \frac{44 \cdot 2}{\pi} = \frac{88}{\pi} $$
Given $$\pi = \frac{22}{7}$$, we substitute this value into the equation:
$$ M_{	ext{new}} = \frac{88 \cdot 7}{22} = 28 \, 	ext{Am}^2 $$
Therefore, the magnetic moment of the strip when bent into a semicircular shape is $$28 \, 	ext{Am}^2$$.

Magnetism and Matter

A straight magnetic strip has a magnetic moment of $44 \mathrm{~Am}^2. If the strip is bent in a semicircular shape, its magnetic moment will be ________ \mathrm{Am}^2. (given \pi=\frac{22}{7}$)

Choose the correct answer: