Magnetism and Matter

For most stable state, $\theta_1=0^{\circ}$ and for most unstable state $\theta_2=180^{\circ}$

∴ Work done $=U_{\text {final }}-U_{\text {initial }}$

$ \begin{aligned} & W=-m B \cos \left(180^{\circ}\right)-\left(-m B \cos 0^{\circ}\right) \\ & W=-m B(-1)+m B=2 m B \\ & W=2 \times 0.5 \times 0.2 \\ & W=0.20 \mathrm{~J} \end{aligned} $

To find the work done in turning the magnet, we need to understand the relationship between the magnetic moment, the magnetic field, and the potential energy in the system. The potential energy (U) of a magnetic dipole in a magnetic field is given by the formula:

$U = -\vec{M} \cdot \vec{B}$

where:

$\vec{M}$ = Magnetic moment vector

$\vec{B}$ = Magnetic field vector

For a bar magnet with a magnetic moment $2M$ in a magnetic field $\frac{B}{2}$, the potential energy can be simplified as:

$U = - (2M) \left( \frac{B}{2} \right) \cos(\theta)$

Initially, let's consider when the magnet is aligned with the magnetic field:

$\theta = 0^\circ$

$U_1 = - (2M) \left( \frac{B}{2} \right) \cos(0^\circ)$

$U_1 = - (2M) \left( \frac{B}{2} \right) \cdot 1$

$U_1 = -MB$

Next, let's consider the state when the magnet is turned by $90^\circ$:

$\theta = 90^\circ$

$U_2 = - (2M) \left( \frac{B}{2} \right) \cos(90^\circ)$

$U_2 = - (2M) \left( \frac{B}{2} \right) \cdot 0$

$U_2 = 0$

The work done (W) is equal to the change in potential energy, which is:

$W = U_1 - U_2$

$W = (-MB) - 0$

$W = MB$

Hence, the work done to turn the magnet by $90^\circ$ in a magnetic field $\frac{B}{2}$ is:

Option D: $MB$

For permanent magnet we prefer a material with high retentivity and high coercivity. For electromagnet, we prefer high saturated magnetism, low coercivity and least possible area of hysteresis loop. Therefore, X is suitable for making permanent magnet and Y for making electromagnet.

Given, area A = 5 cm² = 5 $\times$ 10^$-$4 m²

Number of turns per unit length, n = 4000 turns/m

Magnetic field, B = $\pi$T

Current, i = 5A

Now, magnetic intensity, H = ni = 4000 $\times$ 5 = 2 $\times$ 10⁴ Am^$-$1

As, B = $\mu$₀ (H + I) [I = intensity of magnetisation]

Substituting all the values in above equation, we get

$\pi = 4\pi \times {10^{ - 7}}((2 \times {10^4}) + I)$

${1 \over 4} \times {10^7} = (2 \times {10^4}) + I$

$250 \times {10^4} = 2 \times {10^4} + I$

$I = 250 \times {10^4} - 2 \times {10^4}$

$ \Rightarrow I = 248 \times {10^4}$ A/m

Now, pole strength, $m = I \times A = 248 \times {10^4} \times 5 \times {10^{ - 4}}$

= 1240 A-m