QJEEM24Ph6443: JEE Main 2024 Physics - Class 12 JEEM24Ph6443 Solution: A straight magnetic strip has a magnetic moment of . If the…

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 $44 \, \text{Am}^2$ . 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:

$M_{\text{straight}} = m \cdot l$

where:

• $m$ is the pole strength
• $l$ is the magnetic length

Given $M_{\text{straight}} = 44 \, \text{Am}^2$ , 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, $l_{\text{new}}$ ) is twice the radius:

$l_{\text{new}} = 2R = 2 \cdot \frac{L}{\pi} = \frac{2L}{\pi}$

Now, the new magnetic moment $M_{\text{new}}$ is:

$M_{\text{new}} = m \cdot l_{\text{new}} = m \cdot \frac{2L}{\pi}$

We know from the original magnetic strip:

$M_{\text{straight}} = m \cdot L = 44 \, \text{Am}^2$

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:

$M_{\text{new}} = \frac{44}{L} \cdot \frac{2L}{\pi}$

Canceling out $L$ from the numerator and the denominator:

$M_{\text{new}} = \frac{44 \cdot 2}{\pi} = \frac{88}{\pi}$

Given $\pi = \frac{22}{7}$ , we substitute this value into the equation:

$M_{\text{new}} = \frac{88 \cdot 7}{22} = 28 \, \text{Am}^2$

Therefore, the magnetic moment of the strip when bent into a semicircular shape is $28 \, \text{Am}^2$ .