Class 12 Physics MCQs | Chapter 4: Moving Charges and Magnetism – Part 2

Explanation: The magnetic force on a straight conductor is \(F=IlB\sin\theta\). At \(\theta=0^\circ\), \(\sin0^\circ=0\), so the force is zero, not maximum. At \(\theta=90^\circ\), \(\sin90^\circ=1\), so the force is maximum. At \(\theta=180^\circ\), \(\sin180^\circ=0\), so the force is again zero. At \(\theta=30^\circ\), \(\sin30^\circ=\frac{1}{2}\), so the force is not zero. The same sine-angle logic used for moving charges also appears in the force on current-carrying conductors.

Explanation: \( \textbf{Force relation:} \) \[ F=IlB\sin\theta \] \( \textbf{Fixed quantities:} \) Here \(l\), \(B\), and \(\theta\) are constant. \( \textbf{Linear form:} \) \[ F=(lB\sin\theta)I \] \( \textbf{Graph comparison:} \) This is of the form \(y=mx\), where \(F\) is on the vertical axis and \(I\) is on the horizontal axis. \( \textbf{Slope:} \) The slope is \(lB\sin\theta\). \( \textbf{Unit check:} \) Since slope is \(\frac{F}{I}\), its unit is \(\text{N A}^{-1}\), which matches \(lB\sin\theta\). \( \textbf{Common mistake:} \) The plotted variable \(I\) should not remain inside the slope. \( \textbf{Final answer:} \) The slope is \(lB\sin\theta\).

Explanation: \( \textbf{Given:} \) \(m=20\,\text{g}=0.020\,\text{kg}\), \(l=0.50\,\text{m}\), \(B=0.40\,\text{T}\), \(g=10\,\text{m s}^{-2}\), and \(\theta=90^\circ\). \( \textbf{Required:} \) Current \(I\) for magnetic force to balance weight. \( \textbf{Weight:} \) \[ W=mg=(0.020)(10)=0.20\,\text{N} \] \( \textbf{Magnetic force:} \) \[ F=IlB\sin\theta \] \( \textbf{Perpendicular condition:} \) Since the wire is perpendicular to \(\vec{B}\), \(\sin90^\circ=1\). \( \textbf{Balance condition:} \) \[ IlB=mg \] \( \textbf{Solve for current:} \) \[ I=\frac{mg}{lB} \] \( \textbf{Substitution:} \) \[ I=\frac{0.20}{(0.50)(0.40)} \] \( \textbf{Final simplification:} \) \[ I=\frac{0.20}{0.20}=1.0\,\text{A} \] \( \textbf{Direction check:} \) The current direction must be chosen so that the magnetic force is upward. \( \textbf{Final answer:} \) The required current magnitude is \(1.0\,\text{A}\).

Explanation: The direction of force is found from \(\vec{F}=I(\vec{l}\times\vec{B})\). Take east as \(+x\), north as \(+y\), and vertically upward as \(+z\). The current direction is north, so \(\vec{l}\) is along \(+y\), and \(\vec{B}\) is along \(+z\). The cross product \(+y\times+z=+x\), which is east. The force is not vertically upward because it must be perpendicular to \(\vec{B}\). It is also not along the current because magnetic force is perpendicular to the current direction.

Explanation: The force on a straight current-carrying conductor is \(\vec{F}=I(\vec{l}\times\vec{B})\), so Statement I is correct. Its magnitude is \(F=IlB\sin\theta\). When the conductor is perpendicular to \(\vec{B}\), \(\theta=90^\circ\), and the force is maximum, so Statement II is correct. When the conductor is parallel to \(\vec{B}\), \(\theta=0^\circ\) or \(\theta=180^\circ\), and the force is zero, so Statement III is correct. Statement IV is false because the force is perpendicular to the current direction, not along it. The vector nature of the force is the main vector-direction mistake in magnetic force questions.

Explanation: The force magnitude on each straight portion is \(F=IlB\sin\theta\). Both portions have the same current magnitude \(I\), same length \(l\), same magnetic field \(B\), and both are perpendicular to the vertical field, so the force magnitudes are equal. However, the current directions in the two portions are opposite. Since \(\vec{F}=I(\vec{l}\times\vec{B})\), reversing \(\vec{l}\) reverses the force direction. Therefore, the forces are equal and opposite. This idea becomes important in understanding how a current loop can experience torque even when the net force is zero.

Explanation: In a uniform magnetic field, opposite sides of a current loop can experience equal and opposite magnetic forces. Equal and opposite forces do not necessarily cancel rotational effect if their lines of action are different. Such a pair of forces forms a couple and produces torque. The net translational force can be zero while the loop still tends to rotate. The current is the same throughout a simple loop, so torque is not caused by unequal current in the two sides. The important distinction is between net force, which affects translation, and torque, which affects rotation.

Explanation: A current loop behaves like a magnetic dipole. For a single turn, the magnetic dipole moment magnitude is \(m=IA\). If the coil has \(N\) identical turns, the magnetic moment becomes \(m=NIA\). The magnetic field \(B\) is not part of the magnetic moment itself; it appears in the torque relation \(\tau=mB\sin\theta\). The direction of \(\vec{m}\) is perpendicular to the plane of the loop and is decided by the right-hand rule. The common formula confusion is to include \(B\) in \(\vec{m}\), but \(B\) is the external field acting on the dipole.

Explanation: The torque on a magnetic dipole in a uniform magnetic field is \(\vec{\tau}=\vec{m}\times\vec{B}\). The magnitude is \(\tau=mB\sin\theta\), where \(\theta\) is the angle between \(\vec{m}\) and \(\vec{B}\). The order of the cross product matters because \(\vec{B}\times\vec{m}\) gives the opposite direction. The torque tends to rotate the magnetic dipole moment toward alignment with the magnetic field. The cosine factor belongs to potential energy \(U=-mB\cos\theta\), not to torque magnitude. The angle convention is a frequent mistake: \(\theta\) is measured between \(\vec{m}\) and \(\vec{B}\), not necessarily between the plane of the coil and \(\vec{B}\).

Explanation: \( \textbf{Given:} \) \(N=50\), \(A=2.0\times10^{-3}\,\text{m}^2\), \(I=0.40\,\text{A}\), \(B=0.50\,\text{T}\), and \(\theta=90^\circ\). \( \textbf{Required:} \) Torque \(\tau\) on the coil. \( \textbf{Torque relation:} \) \[ \tau=NIAB\sin\theta \] \( \textbf{Why \(\sin\theta=1\):} \) The magnetic moment \(\vec{m}\) is perpendicular to \(\vec{B}\). \( \textbf{Substitution:} \) \[ \tau=(50)(0.40)(2.0\times10^{-3})(0.50)(1) \] \( \textbf{Intermediate simplification:} \) \[ (50)(0.40)=20 \] \[ (20)(2.0\times10^{-3})=4.0\times10^{-2} \] \( \textbf{Final simplification:} \) \[ \tau=(4.0\times10^{-2})(0.50)=2.0\times10^{-2}\,\text{N m} \] \( \textbf{Unit check:} \) Torque is measured in \(\text{N m}\). \( \textbf{Final answer:} \) The torque is \(2.0\times10^{-2}\,\text{N m}\).

Explanation: The torque on a current loop in a uniform magnetic field is \(\tau=mB\sin\theta\). Here \(\theta\) is the angle between \(\vec{m}\) and \(\vec{B}\), not necessarily the angle between the plane of the loop and \(\vec{B}\). Torque is maximum when \(\sin\theta\) is maximum. Since \(\sin90^\circ=1\), the torque is maximum at \(\theta=90^\circ\). At \(\theta=0^\circ\) or \(\theta=180^\circ\), \(\sin\theta=0\), so the torque is zero. The key angle-related step is to identify the direction of \(\vec{m}\), which is perpendicular to the plane of the loop.

Explanation: \( \textbf{Given:} \) \(N=100\), \(I=0.20\,\text{A}\), \(A=4.0\times10^{-3}\,\text{m}^2\), and \(B=0.50\,\text{T}\). \( \textbf{Important angle condition:} \) The plane of the coil is parallel to \(\vec{B}\). \( \textbf{Magnetic moment direction:} \) \(\vec{m}\) is perpendicular to the plane of the coil. \( \textbf{Angle between \(\vec{m}\) and \(\vec{B}\):} \) If the plane is parallel to \(\vec{B}\), then \(\vec{m}\perp\vec{B}\), so \(\theta=90^\circ\). \( \textbf{Torque formula:} \) \[ \tau=NIAB\sin\theta \] \( \textbf{Substitution:} \) \[ \tau=(100)(0.20)(4.0\times10^{-3})(0.50)\sin90^\circ \] \( \textbf{Simplify:} \) \[ (100)(0.20)=20 \] \[ (20)(4.0\times10^{-3})=8.0\times10^{-2} \] \[ (8.0\times10^{-2})(0.50)=4.0\times10^{-2} \] \( \textbf{Unit check:} \) Torque is measured in \(\text{N m}\). \( \textbf{Final answer:} \) The torque is \(4.0\times10^{-2}\,\text{N m}\).

Explanation: The torque magnitude is \(\tau=mB\sin\theta\). When \(\vec{m}\parallel\vec{B}\), \(\theta=0^\circ\), so \(\tau=0\). The potential energy of a magnetic dipole in a uniform magnetic field is \(U=-mB\cos\theta\). At \(\theta=0^\circ\), \(U=-mB\), which is the minimum value. This is a stable orientation because a small disturbance produces a restoring tendency. The common confusion is to think zero torque always means unstable equilibrium, but the energy test distinguishes stable and unstable cases.

Explanation: When \(\vec{m}\) is antiparallel to \(\vec{B}\), the angle is \(\theta=180^\circ\). The torque magnitude is \(\tau=mB\sin\theta\), so \(\tau=mB\sin180^\circ=0\). However, the potential energy is \(U=-mB\cos180^\circ=+mB\), which is maximum. A maximum-energy orientation is unstable because a slight angular displacement lets the dipole rotate toward lower energy. Thus, zero torque alone does not prove stability. The stability check must be made using potential energy or the effect of a small displacement.

Explanation: The torque is \(\tau=mB\sin\theta\), and the potential energy is \(U=-mB\cos\theta\). At \(\theta=0^\circ\), torque is zero and \(U=-mB\), so row P is wrong because it says torque is maximum. At \(\theta=90^\circ\), torque is maximum and \(U=0\), so row Q is correct. At \(\theta=180^\circ\), torque is zero and \(U=+mB\), so row R is wrong because it says torque is maximum. Row S is also wrong because at \(90^\circ\), torque is not zero. The sine relation controls torque, while the cosine relation controls potential energy.

Explanation: \( \textbf{Given:} \) \(m=0.60\,\text{A m}^2\), \(B=0.25\,\text{T}\), and \(\theta=30^\circ\). \( \textbf{Required:} \) Torque \(\tau\) and potential energy \(U\). \( \textbf{Torque relation:} \) \[ \tau=mB\sin\theta \] \( \textbf{Potential energy relation:} \) \[ U=-mB\cos\theta \] \( \textbf{Common product:} \) \[ mB=(0.60)(0.25)=0.150 \] \( \textbf{Torque calculation:} \) \[ \tau=(0.150)\sin30^\circ=(0.150)\left(\frac{1}{2}\right)=0.075\,\text{N m} \] \( \textbf{Energy calculation:} \) \[ U=-(0.150)\cos30^\circ \] \( \textbf{Use \(\cos30^\circ\approx0.866\):} \) \[ U=-(0.150)(0.866)=-0.1299\,\text{J}\approx-0.130\,\text{J} \] \( \textbf{Sign check:} \) The energy is negative because \(\theta=30^\circ\) is closer to stable alignment than to anti-alignment. \( \textbf{Final answer:} \) \(\tau=0.075\,\text{N m}\) and \(U=-0.130\,\text{J}\).

Explanation: The direction of magnetic dipole moment of a current loop is found by the right-hand rule. Curl the fingers of the right hand in the direction of current in the loop. The extended thumb then gives the direction of \(\vec{m}\). If the current appears anticlockwise to the observer, the thumb points out of the plane toward the observer. The magnetic moment is not tangent to the loop or radial in the plane. The direction of \(\vec{m}\) is perpendicular to the plane of the current loop.

Explanation: The magnetic dipole moment of a current loop has magnitude \(m=IA\) for one turn and \(m=NIA\) for \(N\) turns. Current \(I\) is measured in \(\text{A}\). Area \(A\) is measured in \(\text{m}^2\). Therefore, the unit of \(m\) is \(\text{A m}^2\). The unit \(\text{T}\) belongs to magnetic field, not magnetic moment. The unit-related mistake is to confuse magnetic moment with magnetic field or torque.

Explanation: The magnetic moment of an \(N\)-turn coil is \(m=NIA\), so P matches 1. The torque magnitude on the coil is \(\tau=NIAB\sin\theta\), so Q matches 2. The potential energy of a magnetic dipole in a uniform magnetic field is \(U=-mB\cos\theta\), so R matches 3. The unit of magnetic moment is \(\text{A m}^2\), so S matches 4. The important distinction is that \(B\) appears in torque and energy because an external field is acting, but \(B\) is not part of the magnetic moment \(NIA\).

Explanation: \( \textbf{Given:} \) \(r=7.0\,\text{cm}=0.070\,\text{m}\), \(N=200\), and \(I=0.50\,\text{A}\). \( \textbf{Required:} \) Magnetic dipole moment \(m\). \( \textbf{Area of circular coil:} \) \[ A=\pi r^2 \] \( \textbf{Substitution for area:} \) \[ A=\frac{22}{7}(0.070)^2 \] \( \textbf{Radius square:} \) \[ (0.070)^2=0.0049\,\text{m}^2 \] \( \textbf{Area value:} \) \[ A=\frac{22}{7}(0.0049)=0.0154\,\text{m}^2 \] \( \textbf{Magnetic moment relation:} \) \[ m=NIA \] \( \textbf{Substitution:} \) \[ m=(200)(0.50)(0.0154) \] \( \textbf{Final simplification:} \) \[ m=100(0.0154)=1.54\,\text{A m}^2 \] \( \textbf{Final answer:} \) The magnetic dipole moment is \(1.54\,\text{A m}^2\).