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Q1. Since $B_H = B_V$, $\tan \delta = 1$, so Angle of dip ($\delta$) = $45^\circ$. Q2. High retentivity and High coercivity. Q3. At lower temperatures, thermal agitation is reduced, allowing magnetic dipoles to align more easily with the applied field. Q4. It decreases rapidly and becomes zero at the Curie point (becomes paramagnetic). Q5. The place is at the magnetic poles (Angle of dip = $90^\circ$). Q8. (a) Work Done $W = MB(\cos \theta_1 - \cos \theta_2) = 6 \times 0.44 (\cos 60^\circ - \cos 90^\circ) = 1.32\text J$. (b) Torque $\tau = MB \sin 90^\circ = 6 \times 0.44 \times 1 = 2.64\text Nm$. Q9. Vertical Component $B_V = B_H \tan \delta = 0.3 \tan 60^\circ \approx 0.52\text G$. Resultant $B = B_H / \cos \delta = 0.3 / \cos 60^\circ = 0.6\text G$. Quick interpretation "mr bhatti 57 obb file download
Use Antivirus Software: Ensure your device has reputable antivirus software installed to scan the files for any potential threats. Open your file manager and go to Internal