Moving Charges and Magnetism MCQs with Answers – Part 3 (Class 12 Physics)

Explanation: \( \textbf{Loop field at centre:} \) \[ B_{\text{loop}}=\frac{\mu_0 I}{2R} \] \( \textbf{Long-wire field at distance \(R\):} \) \[ B_{\text{wire}}=\frac{\mu_0 I}{2\pi R} \] \( \textbf{Ratio:} \) \[ \frac{B_{\text{loop}}}{B_{\text{wire}}}=\frac{\mu_0 I/(2R)}{\mu_0 I/(2\pi R)} \] \( \textbf{Cancel common factors:} \) \[ \frac{B_{\text{loop}}}{B_{\text{wire}}}=\pi \] So \(B_{\text{loop}}:B_{\text{wire}}=\pi:1\). The two formulas look similar, but the long-wire result contains the extra factor \(\pi\) in the denominator. \( \textbf{Final answer:} \) The ratio is \(\pi:1\).

Explanation: The magnetic field at the centre of an \(N\)-turn circular coil is \(B=\frac{\mu_0 NI}{2R}\). For fixed \(N\) and \(R\), all factors except \(I\) are constant. Thus \(B\propto I\). If \(I=0\), the magnetic field due to the coil is also zero, so the graph passes through the origin. The straight-line nature shows that doubling current doubles the centre field.

Explanation: \( \textbf{Centre-field relation:} \) \[ B=\frac{\mu_0 I}{2R} \] \( \textbf{First loop:} \) \[ B_1=\frac{\mu_0 I}{2R} \] \( \textbf{Second loop:} \) The current is \(4I\), and the radius is \(2R\). \[ B_2=\frac{\mu_0(4I)}{2(2R)} \] \( \textbf{Simplify:} \) \[ B_2=\frac{4\mu_0 I}{4R}=\frac{\mu_0 I}{R} \] \( \textbf{Compare with \(B_1\):} \) \[ B_1=\frac{\mu_0 I}{2R} \] \[ \frac{B_2}{B_1}=2 \] The current increase is stronger than the radius increase in this comparison. \( \textbf{Final answer:} \) \(B_2:B_1=2:1\).

Explanation: The magnetic field on the axis of a circular loop is \(B=\frac{\mu_0IR^2}{2(R^2+x^2)^{3/2}}\) for a single turn. The direction is along the axis of the loop, with sense decided by the current direction. The expression reduces to the centre-field formula when \(x=0\). The denominator contains \((R^2+x^2)^{3/2}\), which reflects the geometry from every element of the loop to the axial point. This result differs from the long-straight-wire formula because the geometry is circular and axial.

Explanation: The axial distance \(x\) is measured from the centre of the circular loop along its axis. At the centre itself, the observation point is not displaced along the axis, so \(x=0\). Substituting \(x=0\) gives \(B=\frac{\mu_0IR^2}{2(R^2)^{3/2}}\). Since \((R^2)^{3/2}=R^3\), the expression becomes \(B=\frac{\mu_0I}{2R}\). This confirms that the axial-field formula includes the centre-field result as a special case.

Explanation: If all \(N\) turns are closely wound and carry the same current, each turn produces the same magnetic field at the axial point. The contributions are along the same axis and in the same direction. Therefore, the resultant field is \(N\) times the field of a single turn. For one turn, \(B=\frac{\mu_0IR^2}{2(R^2+x^2)^{3/2}}\). For \(N\) turns, it becomes \(B=\frac{\mu_0NIR^2}{2(R^2+x^2)^{3/2}}\).

Explanation: The axial field of a circular loop is \(B=\frac{\mu_0IR^2}{2(R^2+x^2)^{3/2}}\). Far from the loop, \(x\gg R\), so \(R^2+x^2\approx x^2\). Then \((R^2+x^2)^{3/2}\approx(x^2)^{3/2}=x^3\). The numerator \(\mu_0IR^2\) is constant for a fixed loop. Hence the far axial field varies approximately as \(\frac{1}{x^3}\), like the field pattern of a magnetic dipole along its axis.

Explanation: The axial field magnitude is \(B=\frac{\mu_0IR^2}{2(R^2+x^2)^{3/2}}\). The distance variable appears as \(x^2\), so replacing \(x\) by \(-x\) gives the same magnitude. This makes the \(B\)-versus-\(x\) graph symmetric about the centre of the loop. The field is maximum at \(x=0\), not zero. The symmetry comes from the geometry of the circular loop, not from any reversal of current.

Explanation: \( \textbf{Given data:} \) \(R=0.10\,\text{m}\), \(x=0.10\,\text{m}\), \(I=5.0\,\text{A}\), and \(\mu_0=4\pi\times10^{-7}\,\text{T m A}^{-1}\). \( \textbf{Axial-field relation:} \) \[ B=\frac{\mu_0IR^2}{2(R^2+x^2)^{3/2}} \] \( \textbf{Square terms:} \) \[ R^2=x^2=(0.10)^2=0.01 \] \[ R^2+x^2=0.02=2\times10^{-2} \] \( \textbf{Power term:} \) \[ (R^2+x^2)^{3/2}=(2R^2)^{3/2}=2\sqrt{2}R^3 \] because \(x=R\). \( \textbf{Use simplified form for \(x=R\):} \) \[ B=\frac{\mu_0IR^2}{2(2\sqrt{2}R^3)} \] \[ B=\frac{\mu_0 I}{4\sqrt{2}R} \] \( \textbf{Substitution:} \) \[ B=\frac{(4\pi\times10^{-7})(5.0)}{4\sqrt{2}(0.10)} \] \( \textbf{Simplify:} \) \[ B=\frac{20\pi\times10^{-7}}{0.4\sqrt{2}} \] \[ B=\frac{50\pi\times10^{-7}}{\sqrt{2}}=\frac{5\pi\times10^{-6}}{\sqrt{2}} \] This calculated form corresponds to \(\frac{5\pi}{\sqrt{2}}\times10^{-6}\,\text{T}\), so the listed expression matching the substitution is selected. \( \textbf{Final answer:} \) The magnetic field is \(\frac{5\pi}{\sqrt{2}}\times10^{-6}\,\text{T}\).

Explanation: The axial magnetic field of a circular loop is \(B=\frac{\mu_0IR^2}{2(R^2+x^2)^{3/2}}\). For a fixed loop and fixed current, the numerator remains constant. The denominator is smallest when \(x=0\), which corresponds to the centre of the loop. As \(|x|\) increases, \(R^2+x^2\) increases and the magnetic field decreases. The centre is therefore the position of maximum axial field for a given circular loop.

Explanation: \( \textbf{Axial-field relation:} \) \[ B=\frac{\mu_0IR^2}{2(R^2+x^2)^{3/2}} \] \( \textbf{At the centre:} \) \[ B_0=\frac{\mu_0 I}{2R} \] \( \textbf{At \(x=R\):} \) \[ R^2+x^2=R^2+R^2=2R^2 \] \[ (R^2+x^2)^{3/2}=(2R^2)^{3/2}=2\sqrt{2}R^3 \] \( \textbf{Substitution:} \) \[ B=\frac{\mu_0IR^2}{2(2\sqrt{2}R^3)} \] \[ B=\frac{\mu_0 I}{4\sqrt{2}R} \] \( \textbf{Compare with \(B_0\):} \) \[ B_0=\frac{\mu_0 I}{2R} \] \[ B=\frac{B_0}{2\sqrt{2}} \] The field falls because the axial point is farther from every current element than the centre. \( \textbf{Final answer:} \) \(B=\frac{B_0}{2\sqrt{2}}\).

Explanation: The direction of the magnetic field at the centre of a circular loop is obtained by the right-hand rule. Curl the fingers of the right hand in the direction of current. If the current appears clockwise to an observer, the thumb points away from that observer. Hence the axial magnetic field at the centre is into the page from that viewpoint. The field is axial because transverse components from opposite current elements cancel by symmetry.

Explanation: At the centre of a circular loop, the magnetic field is along the axis of the loop, not tangential to the loop. Row P gives the correct centre-field magnitude for a single turn. Row Q gives the standard axial-field expression. Row R follows from the axial formula when \(x\gg R\), because the denominator behaves like \(x^3\). Row S fails because tangential directions around the loop do not survive in the net field at the centre.

Explanation: Ampere’s circuital law states that the closed line integral of magnetic field around an Amperian loop equals \(\mu_0\) times the net current passing through the surface bounded by the loop. The line integral uses the dot product \(\vec{B}\cdot d\vec{l}\), so only the component of \(\vec{B}\) along the path element contributes. The current in the law is the enclosed current, not necessarily every current present nearby. The law is especially useful when symmetry makes \(\vec{B}\) constant in magnitude along a chosen path. Using area integral \(B\,dA\) would describe flux-like reasoning, not Ampere’s circuital law.

Explanation: The magnetic field around a long straight wire forms concentric circles centred on the wire. On a circular Amperian loop centred on the wire, the magnetic field is tangential everywhere and has the same magnitude at all points of the loop. This makes \(\vec{B}\cdot d\vec{l}=B\,dl\) along the path. The closed integral then becomes \(B(2\pi r)\). Choosing a path that does not follow the symmetry makes the integral harder even though the law remains valid.

Explanation: For a long straight wire, a circular Amperian loop of radius \(r\) is chosen around the wire. The magnetic field has constant magnitude on this circle and is tangent to it. Therefore, \(\oint \vec{B}\cdot d\vec{l}=B\oint dl\). The integral \(\oint dl\) around the circle is its circumference \(2\pi r\). This gives \(B(2\pi r)=\mu_0 I\), leading to \(B=\frac{\mu_0 I}{2\pi r}\).

Explanation: \( \textbf{Sign convention:} \) Current out of the page is positive, so current into the page is negative. \( \textbf{Enclosed currents:} \) \[ I_1=+3\,\text{A} \] \[ I_2=-5\,\text{A} \] \( \textbf{Net enclosed current:} \) \[ I_{\text{enc}}=I_1+I_2 \] \[ I_{\text{enc}}=3-5=-2\,\text{A} \] Ampere’s law uses algebraic current enclosed by the loop, not the sum of magnitudes. The negative sign indicates that the net current is into the page under the chosen convention. \( \textbf{Final answer:} \) \(I_{\text{enc}}=-2\,\text{A}\).

Explanation: Ampere’s circuital law involves the net current enclosed by the chosen closed path. If the current-carrying wire lies outside the path, then \(I_{\text{enc}}=0\) for that path. Therefore, \(\oint \vec{B}\cdot d\vec{l}=\mu_0 I_{\text{enc}}=0\). This does not mean the magnetic field is zero at every point on the path. It means that the signed line integral around that closed path is zero because no net current pierces the enclosed surface.

Explanation: Ampere’s circuital law is written for a closed path, so statement I is true. The current in the law is the algebraic enclosed current, with signs decided by the chosen orientation of the loop, so statement II is also true. Statement III is false because a line integral can be zero even when the magnetic field is non-zero at different points on the path. Contributions to the integral may cancel, or no current may be enclosed. A zero closed integral is not the same as zero magnetic field everywhere.

Explanation: Inside a long ideal solenoid, the magnetic field is nearly uniform and directed along the axis. The standard expression is \(B=\mu_0 nI\), where \(n\) is the number of turns per unit length and \(I\) is the current. The long-wire expression \(\frac{\mu_0 I}{2\pi r}\) applies around a straight wire, not inside a solenoid. The circular-loop centre formula applies to a single loop or coil centre, not to the uniform interior field of a long solenoid. The factor \(n\) shows that more turns per unit length strengthen the solenoid field.