Magnetism and Matter MCQs with Answers – Part 2 (Class 12 Physics)

Explanation: \( \textbf{Torque relation:} \) \[ \tau=mB\sin\theta \] In the graph, the vertical quantity is \(\tau\), and the horizontal quantity is \(\sin\theta\). Rewrite the relation as: \[ \tau=(mB)(\sin\theta) \] This has the straight-line form \(y=(\text{slope})x\). Therefore the slope is \(mB\). The slope stays constant only because \(m\) and \(B\) are fixed during the graph. \( \textbf{Final answer:} \) The slope of the \(\tau\) versus \(\sin\theta\) graph is \(mB\). The expression \(mB\cos\theta\) belongs to energy-related reasoning, not to this torque graph slope.

Explanation: \( \textbf{Given:} \) \(m=2.0\,A\,m^2\), \(\tau=0.60\,N\,m\), and \(\theta=90^\circ\). \( \textbf{Required:} \) Magnetic field magnitude \(B\). \( \textbf{Torque formula:} \) \[ \tau=mB\sin\theta \] At \(\theta=90^\circ\), \(\sin90^\circ=1\). So the formula becomes: \[ \tau=mB \] \( \textbf{Solving for \(B\):} \) \[ B=\frac{\tau}{m} \] \( \textbf{Substitution:} \) \[ B=\frac{0.60}{2.0} \] \[ B=0.30\,T \] \( \textbf{Final answer:} \) The magnetic field magnitude is \(0.30\,T\). The unit \(T\) is consistent because torque divided by magnetic dipole moment gives magnetic field.

Explanation: A magnetic dipole in a uniform magnetic field can have equal and opposite forces acting at different points of the dipole. Equal and opposite forces give zero net translational force. However, if their lines of action are not the same, they form a couple and produce torque. This is why the dipole may rotate even though it does not accelerate as a whole in a uniform field. The Reason directly explains how torque can exist without net force.

Explanation: The torque vector is given by the cross product \(\vec{\tau}=\vec{m}\times\vec{B}\). A cross product gives a vector perpendicular to the plane containing the two vectors being crossed. Its direction is decided by the right-hand rule applied from \(\vec{m}\) toward \(\vec{B}\). The torque vector is therefore not generally along \(\vec{m}\) or along \(\vec{B}\). The magnitude \(mB\sin\theta\) gives the size of the turning effect, while the vector direction gives the axis sense of rotation.

Explanation: \( \textbf{Given directions:} \) \(\vec{m}\) is along \(+x\), and \(\vec{B}\) is along \(+y\). \( \textbf{Torque relation:} \) \[ \vec{\tau}=\vec{m}\times\vec{B} \] So the direction is the same as: \[ (+x)\times(+y) \] Using the right-hand rule: \[ \hat{i}\times\hat{j}=\hat{k} \] Therefore the torque is along \(+z\). \( \textbf{Final answer:} \) The torque direction is \(+z\). Reversing the order to \(\vec{B}\times\vec{m}\) would give the opposite direction, but the torque formula uses \(\vec{m}\times\vec{B}\).

Explanation: When \(\vec{m}\) is parallel to \(\vec{B}\), the torque magnitude is zero because \(\theta=0^\circ\). A small angular displacement produces a restoring tendency that brings the dipole back toward alignment with the field. This makes the parallel orientation a stable equilibrium. It is not the maximum-torque orientation, because torque is maximum at \(\theta=90^\circ\). Stability depends on how the system responds to a small disturbance, not only on the torque being zero at the exact position.

Explanation: At \(\theta=180^\circ\), the torque magnitude is \(\tau=mB\sin180^\circ=0\). However, zero torque at an exact position does not automatically mean stable equilibrium. If the dipole is displaced slightly from the antiparallel orientation, the magnetic torque tends to rotate it away from that position and toward parallel alignment. Therefore the antiparallel orientation is unstable. The contrast with \(\theta=0^\circ\) is based on response to disturbance, not on the value of torque at the exact angle alone.

Explanation: \( \textbf{Torque relation:} \) \[ \tau=mB\sin\theta \] The values of \(m\) and \(B\) are the same in both cases, so only \(\sin\theta\) changes. \( \textbf{At \(30^\circ\):} \) \[ \tau_{30^\circ}=mB\sin30^\circ=mB\left(\frac{1}{2}\right) \] \( \textbf{At \(90^\circ\):} \) \[ \tau_{90^\circ}=mB\sin90^\circ=mB(1) \] \( \textbf{Ratio:} \) \[ \frac{\tau_{30^\circ}}{\tau_{90^\circ}}=\frac{1/2}{1} \] \[ \frac{\tau_{30^\circ}}{\tau_{90^\circ}}=\frac{1}{2} \] \( \textbf{Final answer:} \) \(\tau_{30^\circ}=\frac{1}{2}\tau_{90^\circ}\). The actual values of \(m\) and \(B\) are not needed for this comparison because they cancel in the ratio.

Explanation: The potential energy of a magnetic dipole in a uniform magnetic field is \(U=-\vec{m}\cdot\vec{B}\). In scalar form, this becomes \(U=-mB\cos\theta\). The dot product is used because potential energy is a scalar quantity. The cross product \(\vec{m}\times\vec{B}\) gives torque, not potential energy. The negative sign is important because the parallel orientation has the lowest energy.

Explanation: The potential energy of a magnetic dipole is \(U=-mB\cos\theta\). At \(\theta=0^\circ\), \(\cos0^\circ=1\), so \(U=-mB\). This is the minimum value because it is the most negative energy. At \(\theta=180^\circ\), \(\cos180^\circ=-1\), so \(U=+mB\), which is maximum. The minimum-energy orientation corresponds to stable equilibrium, where \(\vec{m}\) is aligned with \(\vec{B}\).

Explanation: \( \textbf{Given:} \) \(m=0.80\,A\,m^2\), \(B=0.50\,T\), and \(\theta=60^\circ\). \( \textbf{Required:} \) Potential energy \(U\). \( \textbf{Energy relation:} \) \[ U=-mB\cos\theta \] This relation applies to a magnetic dipole in a uniform magnetic field. \( \textbf{Substitution:} \) \[ U=-(0.80)(0.50)\cos60^\circ \] \( \textbf{Trigonometric value:} \) \[ \cos60^\circ=\frac{1}{2} \] \( \textbf{Calculation:} \) \[ U=-(0.80)(0.50)\left(\frac{1}{2}\right) \] \[ U=-(0.40)(0.50) \] \[ U=-0.20\,J \] \( \textbf{Final answer:} \) The potential energy is \(-0.20\,J\). The negative sign shows that this orientation is closer to stable alignment than to the antiparallel orientation.

Explanation: \( \textbf{Potential energy relation:} \) \[ U=-mB\cos\theta \] \( \textbf{Initial angle:} \) \(\theta_i=0^\circ\). \[ U_i=-mB\cos0^\circ=-mB \] \( \textbf{Final angle:} \) \(\theta_f=180^\circ\). \[ U_f=-mB\cos180^\circ \] Since \(\cos180^\circ=-1\): \[ U_f=+mB \] \( \textbf{Change in energy:} \) \[ \Delta U=U_f-U_i \] \[ \Delta U=mB-(-mB) \] \[ \Delta U=2mB \] \( \textbf{Final answer:} \) The potential energy increases by \(2mB\). The rotation is from the lowest-energy orientation to the highest-energy orientation, so the change is positive.

Explanation: The potential energy of a magnetic dipole in a uniform magnetic field is \(U=-mB\cos\theta\). At \(\theta=0^\circ\), \(\cos0^\circ=1\), so \(U=-mB\), which is the minimum value. At \(\theta=90^\circ\), \(\cos90^\circ=0\), so \(U=0\). At \(\theta=180^\circ\), \(\cos180^\circ=-1\), so \(U=+mB\), which is the maximum value. The signs are not arbitrary; they show whether the dipole is aligned with or opposed to the field.

Explanation: \( \textbf{Given:} \) \(m=1.2\,A\,m^2\), \(B=0.50\,T\), \(\theta_i=60^\circ\), and \(\theta_f=120^\circ\). \( \textbf{Required:} \) Work done by an external agent in slow rotation. For slow rotation, external work equals increase in potential energy: \[ W_{ext}=\Delta U=U_f-U_i \] Potential energy of a magnetic dipole: \[ U=-mB\cos\theta \] \( \textbf{Initial energy:} \) \[ U_i=-(1.2)(0.50)\cos60^\circ \] \[ U_i=-0.60\left(\frac{1}{2}\right)=-0.30\,J \] \( \textbf{Final energy:} \) \[ U_f=-(1.2)(0.50)\cos120^\circ \] Since \(\cos120^\circ=-\frac{1}{2}\): \[ U_f=-0.60\left(-\frac{1}{2}\right)=+0.30\,J \] \( \textbf{Change in energy:} \) \[ \Delta U=0.30-(-0.30) \] \[ \Delta U=0.60\,J \] \( \textbf{Final answer:} \) The work done by the external agent is \(0.60\,J\). The positive value shows that the dipole is being moved toward a higher-energy orientation.

Explanation: The graph starts at \(-mB\) when \(\theta=0^\circ\), so the relation must give a negative value at \(\cos0^\circ=1\). The expression \(U=-mB\cos\theta\) gives \(U=-mB\) at \(0^\circ\). At \(\theta=90^\circ\), \(\cos90^\circ=0\), so it gives \(U=0\). At \(\theta=180^\circ\), \(\cos180^\circ=-1\), so it gives \(U=+mB\). A sine relation would give \(0\) at both \(0^\circ\) and \(180^\circ\), which does not match the described graph.

Explanation: Stable equilibrium corresponds to the minimum potential energy position. For a magnetic dipole, \(U=-mB\cos\theta\), and the minimum value occurs at \(\theta=0^\circ\). This means \(\vec{m}\) is parallel to \(\vec{B}\). If the dipole is displaced slightly from this orientation, the magnetic torque tends to restore it toward alignment. The antiparallel position has maximum potential energy, so it is not the stable orientation.

Explanation: Torque magnitude is \(\tau=mB\sin\theta\), so at \(\theta=90^\circ\), torque is maximum. Potential energy is \(U=-mB\cos\theta\), so at \(\theta=90^\circ\), the energy is \(0\). Statement III is partly wrong because at \(\theta=0^\circ\), \(\sin0^\circ=0\), so torque is zero, not maximum. The potential energy at \(\theta=0^\circ\) is indeed minimum. Torque and energy must be judged from different trigonometric factors, \(\sin\theta\) and \(\cos\theta\).

Explanation: \( \textbf{Potential energy relation:} \) \[ U=-mB\cos\theta \] \( \textbf{Initial angle:} \) \(\theta_i=180^\circ\). \[ U_i=-mB\cos180^\circ \] \[ U_i=-mB(-1)=+mB \] \( \textbf{Final angle:} \) \(\theta_f=0^\circ\). \[ U_f=-mB\cos0^\circ \] \[ U_f=-mB(1)=-mB \] \( \textbf{Change in potential energy:} \) \[ \Delta U=U_f-U_i \] \[ \Delta U=-mB-(+mB) \] \[ \Delta U=-2mB \] \( \textbf{Final answer:} \) The potential energy changes by \(-2mB\). The negative sign indicates that the dipole moves from the highest-energy orientation to the lowest-energy orientation.

Explanation: A current loop produces a magnetic field around it. At large distances, the field pattern of a small current loop resembles the field of a magnetic dipole. This is why a current loop is assigned a magnetic dipole moment. The loop does not behave like an isolated magnetic pole because ordinary magnetic field lines remain closed. The dipole description is especially useful when the loop is small compared with the observation distance.

Explanation: The magnetic dipole moment of a plane current loop is given by \(m=IA\). Here \(I\) is the current in the loop and \(A\) is the area enclosed by the loop. A larger current or a larger area produces a larger magnetic dipole moment. The expression is a product, not a sum, because current and area together determine the loop’s magnetic strength. The unit becomes \(A\,m^2\), matching the unit of magnetic dipole moment of a bar magnet.