Magnetism and Matter MCQs with Answers – Part 5 (Class 12 Physics)

Explanation: \( \textbf{First find Earth's horizontal component:} \) \[ B_H=B_E\cos I \] \[ B_H=(50\,\mu T)\cos60^\circ \] \[ B_H=50\times\frac{1}{2}\,\mu T \] \[ B_H=25\,\mu T=25\times10^{-6}\,T \] \( \textbf{Neutral point condition:} \) Since \(\vec{m}\parallel\vec{B}_H\), cancellation occurs on the equatorial line. \[ \frac{\mu_0}{4\pi}\frac{m}{r^3}=B_H \] Solving for \(r^3\): \[ r^3=\frac{\frac{\mu_0}{4\pi}m}{B_H} \] \[ r^3=\frac{(10^{-7})(1.25)}{25\times10^{-6}} \] \[ r^3=\frac{1.25\times10^{-7}}{2.5\times10^{-5}} \] \[ r^3=5.0\times10^{-3}\,m^3 \] \[ r=\sqrt[3]{5.0\times10^{-3}}\,m \] \[ r\approx0.171\,m \] \( \textbf{Final answer:} \) The equatorial neutral point is about \(0.171\,m\) from the centre. The dip angle must be used first because neutral points in the horizontal plane are decided by \(B_H\), not by the total field \(B_E\).

Explanation: \( \textbf{Axial field at P:} \) \[ B_P=\frac{\mu_0}{4\pi}\frac{2m}{(0.20)^3} \] \( \textbf{Equatorial field at Q:} \) \[ B_Q=\frac{\mu_0}{4\pi}\frac{m}{(0.10)^3} \] \( \textbf{Ratio:} \) \[ \frac{B_Q}{B_P}=\frac{\frac{\mu_0}{4\pi}\frac{m}{(0.10)^3}}{\frac{\mu_0}{4\pi}\frac{2m}{(0.20)^3}} \] Cancel common factors: \[ \frac{B_Q}{B_P}=\frac{(0.20)^3}{2(0.10)^3} \] \[ \frac{B_Q}{B_P}=\frac{0.008}{2(0.001)} \] \[ \frac{B_Q}{B_P}=4 \] On the equatorial line: \[ \vec{B}_Q\ \text{is opposite to}\ \vec{m} \] \( \textbf{Final answer:} \) \(\frac{B_Q}{B_P}=4\), and \(\vec{B}_Q\) is opposite to \(\vec{m}\). The equatorial point is closer, so its field can be larger even though the axial formula has the factor \(2\).

Explanation: \( \textbf{Given:} \) \(N=150\), \(r=4.0\,cm=0.040\,m\), \(I=0.50\,A\), and \(B=0.20\,T\). \( \textbf{Area of one turn:} \) \[ A=\pi r^2 \] \[ A=\pi(0.040)^2 \] \[ A=0.0016\pi\,m^2 \] \( \textbf{Magnetic moment of the coil:} \) \[ m=NIA \] \[ m=(150)(0.50)(0.0016\pi) \] \[ m=0.12\pi\,A\,m^2 \] \( \textbf{Potential energy:} \) \[ U=-mB\cos\theta \] At \(\theta_i=0^\circ\): \[ U_i=-mB \] At \(\theta_f=120^\circ\): \[ U_f=-mB\cos120^\circ \] \[ \cos120^\circ=-\frac{1}{2} \] \[ U_f=+\frac{mB}{2} \] \( \textbf{External work in slow rotation:} \) \[ W_{ext}=U_f-U_i \] \[ W_{ext}=\frac{mB}{2}-(-mB) \] \[ W_{ext}=\frac{3mB}{2} \] \[ mB=(0.12\pi)(0.20)=0.024\pi\,J \] \[ W_{ext}=\frac{3}{2}(0.024\pi) \] \[ W_{ext}=0.036\pi\,J \] \( \textbf{Final answer:} \) The work done by the external agent is \(0.036\pi\,J\). The factor \(\frac{3}{2}\) comes from moving from the lowest-energy direction to an angle where the energy is \(+\frac{mB}{2}\).

Explanation: \( \textbf{Magnetisation:} \) \[ M=\chi_mH \] \[ M=(0.25)(1.6\times10^3) \] \[ M=4.0\times10^2\,A\,m^{-1} \] \( \textbf{Net magnetic moment:} \) \[ M=\frac{m_{net}}{V} \] \[ m_{net}=MV \] \[ m_{net}=(4.0\times10^2)(3.0\times10^{-5}) \] \[ m_{net}=1.2\times10^{-2}\,A\,m^2 \] \( \textbf{Magnetic field inside:} \) \[ B=\mu_0(H+M) \] \[ H+M=1.6\times10^3+4.0\times10^2 \] \[ H+M=2.0\times10^3\,A\,m^{-1} \] \[ B=(4\pi\times10^{-7})(2.0\times10^3) \] \[ B=8\pi\times10^{-4}\,T \] \( \textbf{Final answer:} \) \(m_{net}=1.2\times10^{-2}\,A\,m^2\) and \(B=8\pi\times10^{-4}\,T\). The solution needs both the material-response relation \(M=\chi_mH\) and the field relation \(B=\mu_0(H+M)\).

Explanation: \( \textbf{Given at \(300\,K\):} \) \[ \mu_{r1}=1.004 \] \[ \chi_1=\mu_{r1}-1 \] \[ \chi_1=0.004 \] \( \textbf{Curie's law:} \) \[ \chi_m=\frac{C}{T} \] For the same material: \[ \chi_1T_1=\chi_2T_2 \] \[ \chi_2=\chi_1\frac{T_1}{T_2} \] \[ \chi_2=0.004\frac{300}{600} \] \[ \chi_2=0.002 \] Now convert back to relative permeability: \[ \mu_{r2}=1+\chi_2 \] \[ \mu_{r2}=1.002 \] \( \textbf{Final answer:} \) The relative permeability at \(600\,K\) is \(1.002\). The permeability moves closer to \(1\) because the paramagnetic susceptibility decreases as temperature increases.

Explanation: \( \textbf{Initial loss per unit volume per cycle:} \) \[ 75\,J\,m^{-3} \] \( \textbf{Reduction in loop area:} \) \[ 40\%\ \text{of}\ 75=0.40\times75 \] \[ =30\,J\,m^{-3} \] This is the energy saving per unit volume per cycle. \( \textbf{Core volume:} \) \[ V=8.0\times10^{-4}\,m^3 \] Energy saved per cycle is: \[ \Delta E=(30)(8.0\times10^{-4}) \] \[ \Delta E=2.4\times10^{-2}\,J \] At \(50\,Hz\): \[ \Delta P=\Delta E f \] \[ \Delta P=(2.4\times10^{-2})(50) \] \[ \Delta P=1.20\,W \] \( \textbf{Final answer:} \) The power loss is reduced by \(1.20\,W\). The percentage reduction must first be applied to the loop-area loss, then multiplied by volume and frequency.

Explanation: \( \textbf{Axial field dependence:} \) \[ B_{axial}\propto \frac{m}{r^3} \] For the first dipole: \[ B\propto \frac{m}{r^3} \] For the second dipole: \[ B'\propto \frac{4m}{R^3} \] We need \(B'=B\), so: \[ \frac{4m}{R^3}=\frac{m}{r^3} \] Cancel \(m\): \[ \frac{4}{R^3}=\frac{1}{r^3} \] \[ R^3=4r^3 \] \[ R=\sqrt[3]{4}\,r \] \( \textbf{Final answer:} \) The required distance is \(\sqrt[3]{4}\,r\). A fourfold increase in dipole moment requires only a cube-root increase in distance because the dipole field varies as \(\frac{1}{r^3}\).

Explanation: \( \textbf{Horizontal component of Earth's field:} \) \[ B_H=B_E\cos I \] \[ B_H=(60\,\mu T)\cos60^\circ \] \[ B_H=30\,\mu T=30\times10^{-6}\,T \] \( \textbf{Orientation decision:} \) Since \(\vec{m}\) is opposite to \(\vec{B}_H\), neutral points are on the axial line. \( \textbf{Axial neutral condition:} \) \[ \frac{\mu_0}{4\pi}\frac{2m}{r^3}=B_H \] \[ r^3=\frac{(10^{-7})(2)(0.90)}{30\times10^{-6}} \] \[ r^3=\frac{1.8\times10^{-7}}{3.0\times10^{-5}} \] \[ r^3=6.0\times10^{-3}\,m^3 \] \[ r=\sqrt[3]{6.0\times10^{-3}}\,m \] \[ r\approx0.182\,m \] \( \textbf{Final answer:} \) The axial neutral-point distance is approximately \(0.182\,m\). The total field \(B_E\) must first be resolved because the horizontal component is what cancels the magnet's horizontal dipole field.

Explanation: \( \textbf{Area of the rectangular coil:} \) \[ A=(0.080)(0.050) \] \[ A=4.0\times10^{-3}\,m^2 \] \( \textbf{Magnetic moment:} \) \[ m=NIA \] \[ m=(80)(0.25)(4.0\times10^{-3}) \] \[ m=0.080\,A\,m^2 \] \( \textbf{Maximum torque:} \) \[ \tau_{max}=mB \] \[ \tau_{max}=(0.080)(0.50) \] \[ \tau_{max}=0.040\,N\,m \] \( \textbf{Work from stable to unstable equilibrium:} \) Stable equilibrium is \(\theta=0^\circ\): \[ U_i=-mB \] Unstable equilibrium is \(\theta=180^\circ\): \[ U_f=+mB \] \[ W_{ext}=U_f-U_i \] \[ W_{ext}=mB-(-mB) \] \[ W_{ext}=2mB \] \[ W_{ext}=2(0.080)(0.50) \] \[ W_{ext}=0.080\,J \] \( \textbf{Final answer:} \) \(\tau_{max}=0.040\,N\,m\) and \(W_{ext}=0.080\,J\). The torque result uses the maximum value of \(\sin\theta\), while the work result uses the energy difference between two equilibrium orientations.

Explanation: \( \textbf{Relation between \(B\), \(H\), and permeability:} \) \[ B=\mu H \] So: \[ \mu=\frac{B}{H} \] \[ \mu=\frac{3.0\times10^{-3}}{1.0\times10^3} \] \[ \mu=3.0\times10^{-6}\,T\,m\,A^{-1} \] \( \textbf{Relative permeability:} \) \[ \mu_r=\frac{\mu}{\mu_0} \] \[ \mu_r=\frac{3.0\times10^{-6}}{4\pi\times10^{-7}} \] \[ \mu_r=\frac{3.0}{4\pi}\times10 \] \[ \mu_r=\frac{30}{4\pi} \] \[ \mu_r=\frac{7.5}{\pi} \] For a linear magnetic material: \[ \mu_r=1+\chi_m \] \[ \chi_m=\mu_r-1 \] \[ \chi_m=\frac{7.5}{\pi}-1 \] \( \textbf{Final answer:} \) The approximate values are \(\mu_r=\frac{7.5}{\pi}\) and \(\chi_m=\frac{7.5}{\pi}-1\). The susceptibility is found only after first converting the measured \(B/H\) ratio into relative permeability.

Explanation: \( \textbf{Earth's horizontal component:} \) \[ B_H=B_E\cos I \] \[ B_H=52\,\mu T\cos30^\circ \] \[ B_H=52\times\frac{\sqrt{3}}{2}\,\mu T \] \[ B_H\approx45.0\,\mu T=4.50\times10^{-5}\,T \] \( \textbf{Orientation decision:} \) Since \(\vec{m}\) is opposite to \(\vec{B}_H\), cancellation occurs on the axial line. \( \textbf{Axial neutral condition:} \) \[ \frac{\mu_0}{4\pi}\frac{2m}{r^3}=B_H \] \[ r^3=\frac{(10^{-7})(2)(1.60)}{4.50\times10^{-5}} \] \[ r^3=\frac{3.20\times10^{-7}}{4.50\times10^{-5}} \] \[ r^3\approx7.11\times10^{-3}\,m^3 \] \[ r\approx\sqrt[3]{7.11\times10^{-3}}\,m \] \[ r\approx0.192\,m \] \( \textbf{Final answer:} \) The axial neutral-point distance is approximately \(0.192\,m\). The total Earth field cannot be used directly because the neutral-point condition uses the horizontal component \(B_H\).

Explanation: \( \textbf{Area of one circular turn:} \) \[ A=\pi r^2 \] \[ A=\pi(0.030)^2 \] \[ A=9.0\times10^{-4}\pi\,m^2 \] \( \textbf{Magnetic moment of the coil:} \) \[ m=NIA \] \[ m=(250)(0.80)(9.0\times10^{-4}\pi) \] \[ m=0.180\pi\,A\,m^2 \] \( \textbf{Maximum torque:} \) \[ \tau_{max}=mB \] \[ \tau_{max}=(0.180\pi)(0.50) \] \[ \tau_{max}=0.090\pi\,N\,m \] \( \textbf{Work from \(\theta=0^\circ\) to \(\theta=90^\circ\):} \) \[ U=-mB\cos\theta \] \[ U_i=-mB\cos0^\circ=-mB \] \[ U_f=-mB\cos90^\circ=0 \] \[ W_{ext}=U_f-U_i \] \[ W_{ext}=0-(-mB)=mB \] \[ W_{ext}=0.090\pi\,J \] \( \textbf{Final answer:} \) \(\tau_{max}=0.090\pi\,N\,m\) and \(W_{ext}=0.090\pi\,J\). The same product \(mB\) appears in both results here, but it represents maximum torque in one case and an energy change in the other.

Explanation: \( \textbf{Axial field at \(0.30\,m\):} \) \[ B_0=\frac{\mu_0}{4\pi}\frac{2m}{(0.30)^3} \] \( \textbf{Equatorial field at \(0.15\,m\):} \) \[ B_e=\frac{\mu_0}{4\pi}\frac{m}{(0.15)^3} \] \( \textbf{Ratio:} \) \[ \frac{B_e}{B_0}=\frac{\frac{\mu_0}{4\pi}\frac{m}{(0.15)^3}}{\frac{\mu_0}{4\pi}\frac{2m}{(0.30)^3}} \] Cancel common factors: \[ \frac{B_e}{B_0}=\frac{(0.30)^3}{2(0.15)^3} \] Since: \[ 0.30=2(0.15) \] \[ \frac{B_e}{B_0}=\frac{(2)^3}{2} \] \[ \frac{B_e}{B_0}=\frac{8}{2}=4 \] \( \textbf{Final answer:} \) The equatorial field magnitude is \(4B_0\). The closer equatorial point overcomes the missing axial factor \(2\) because distance enters as a cube.

Explanation: \( \textbf{Magnetisation:} \) \[ M=\chi_mH \] \[ M=(0.40)(2.5\times10^3) \] \[ M=1.0\times10^3\,A\,m^{-1} \] \( \textbf{Net magnetic moment:} \) \[ m_{net}=MV \] \[ m_{net}=(1.0\times10^3)(6.0\times10^{-5}) \] \[ m_{net}=6.0\times10^{-2}\,A\,m^2 \] \( \textbf{Magnetic field inside:} \) \[ B=\mu_0(H+M) \] \[ H+M=2.5\times10^3+1.0\times10^3 \] \[ H+M=3.5\times10^3\,A\,m^{-1} \] \[ B=(4\pi\times10^{-7})(3.5\times10^3) \] \[ B=14\pi\times10^{-4}\,T \] \( \textbf{Final answer:} \) \(m_{net}=6.0\times10^{-2}\,A\,m^2\) and \(B=14\pi\times10^{-4}\,T\). The magnetic field inside depends on both the applied magnetic intensity and the material magnetisation.

Explanation: \( \textbf{Initial magnetisation:} \) \[ M_1=\chi_1H \] \( \textbf{Curie's law:} \) \[ \chi_m\propto\frac{1}{T} \] When temperature changes from \(300\,K\) to \(600\,K\): \[ \chi_2=\chi_1\frac{300}{600} \] \[ \chi_2=\frac{\chi_1}{2} \] The magnetising field is doubled: \[ H_2=2H \] \( \textbf{New magnetisation:} \) \[ M_2=\chi_2H_2 \] \[ M_2=\left(\frac{\chi_1}{2}\right)(2H) \] \[ M_2=\chi_1H \] \[ M_2=M_1 \] \( \textbf{Final answer:} \) The new magnetisation is the same as the original magnetisation. The reduced susceptibility is exactly balanced by the doubled magnetising field.

Explanation: \( \textbf{Hysteresis power loss relation:} \) \[ P=(\text{loop area per unit volume per cycle})\times V\times f \] \( \textbf{Old power:} \) \[ P_1=40\times V\times60 \] \( \textbf{New power:} \) \[ P_2=24\times V\times50 \] \( \textbf{Ratio:} \) \[ \frac{P_2}{P_1}=\frac{24\times V\times50}{40\times V\times60} \] Cancel \(V\): \[ \frac{P_2}{P_1}=\frac{24}{40}\times\frac{50}{60} \] \[ \frac{P_2}{P_1}=\frac{3}{5}\times\frac{5}{6} \] \[ \frac{P_2}{P_1}=\frac{1}{2} \] \( \textbf{Final answer:} \) The new power loss is \(\frac{1}{2}\) of the old power loss. Both loop area and cycling frequency must be included because energy is lost once in every cycle.

Explanation: \( \textbf{Magnetic moment from pole model:} \) \[ m=p_m(2l) \] \[ 2l=8.0\,cm=0.080\,m \] \[ m=(10)(0.080) \] \[ m=0.80\,A\,m^2 \] \( \textbf{Orientation decision:} \) Since \(\vec{m}\parallel\vec{B}_H\), the neutral points lie on the equatorial line. \( \textbf{Equatorial neutral condition:} \) \[ \frac{\mu_0}{4\pi}\frac{m}{r^3}=B_H \] \[ r^3=\frac{(10^{-7})(0.80)}{4.0\times10^{-5}} \] \[ r^3=2.0\times10^{-3}\,m^3 \] \[ r=\sqrt[3]{2.0\times10^{-3}}\,m \] \[ r\approx0.126\,m \] \( \textbf{Final answer:} \) The equatorial neutral distance is approximately \(0.126\,m\). The magnetic length must be converted into metres before using it to find the dipole moment.

Explanation: \( \textbf{Given:} \) \(m=0.40\,A\,m^2\), \(B=0.50\,T\), \(\theta_i=45^\circ\), and \(\theta_f=135^\circ\). \( \textbf{Potential energy relation:} \) \[ U=-mB\cos\theta \] \( \textbf{Initial energy:} \) \[ U_i=-mB\cos45^\circ \] \[ U_i=-mB\left(\frac{1}{\sqrt{2}}\right) \] \( \textbf{Final energy:} \) \[ U_f=-mB\cos135^\circ \] \[ \cos135^\circ=-\frac{1}{\sqrt{2}} \] \[ U_f=+\frac{mB}{\sqrt{2}} \] \( \textbf{External work:} \) \[ W_{ext}=U_f-U_i \] \[ W_{ext}=\frac{mB}{\sqrt{2}}-\left(-\frac{mB}{\sqrt{2}}\right) \] \[ W_{ext}=\frac{2mB}{\sqrt{2}} \] \[ W_{ext}=\sqrt{2}mB \] Since: \[ mB=(0.40)(0.50)=0.20\,J \] \[ W_{ext}=0.20\sqrt{2}\,J \] \( \textbf{Final answer:} \) The external work done is \(0.20\sqrt{2}\,J\). The work is positive because the final orientation has higher potential energy than the initial orientation.

Explanation: \( \textbf{Earth field components:} \) \[ \tan I=\frac{B_V}{B_H} \] \[ \frac{B_V}{36}=\frac{3}{4} \] \[ B_V=27\,\mu T \] \( \textbf{Total Earth field:} \) \[ B_E=\sqrt{B_H^2+B_V^2} \] \[ B_E=\sqrt{36^2+27^2}\,\mu T \] \[ B_E=\sqrt{1296+729}\,\mu T \] \[ B_E=\sqrt{2025}\,\mu T \] \[ B_E=45\,\mu T \] \( \textbf{Neutral point condition:} \) Since \(\vec{m}\) is opposite to \(\vec{B}_H\), the neutral point is axial. \[ \frac{\mu_0}{4\pi}\frac{2m}{r^3}=B_H \] \[ r^3=\frac{(10^{-7})(2)(1.0)}{36\times10^{-6}} \] \[ r^3=5.56\times10^{-3}\,m^3 \] \[ r\approx0.177\,m \] \( \textbf{Final answer:} \) \(B_E=45\,\mu T\) and the axial neutral distance is approximately \(0.177\,m\). The total field is found from components, but the neutral distance uses the horizontal component \(B_H\).

Explanation: \( \textbf{Magnetisation:} \) \[ M=\chi_mH \] \[ M=(-2.0\times10^{-4})(3.0\times10^3) \] \[ M=-6.0\times10^{-1}\,A\,m^{-1} \] \[ M=-0.60\,A\,m^{-1} \] \( \textbf{Net magnetic moment:} \) \[ m_{net}=MV \] \[ m_{net}=(-0.60)(2.5\times10^{-4}) \] \[ m_{net}=-1.5\times10^{-4}\,A\,m^2 \] \( \textbf{Relative permeability:} \) \[ \mu_r=1+\chi_m \] \[ \mu_r=1-2.0\times10^{-4} \] \[ \mu_r=0.9998 \] \( \textbf{Final answer:} \) \(M=-0.60\,A\,m^{-1}\), \(m_{net}=-1.5\times10^{-4}\,A\,m^2\), and \(\mu_r=0.9998\). The negative signs show that the material's response is opposite to the applied magnetising field.