CBSE Class 12 Physics: 2025 board paper, set 55/1/1

1. $F=\frac{k{q}_a{q}_b}{r^2}$, so the slope of the $F$–$\frac{1}{r^2}$ line is $k{q}_a{q}_b$. The $(q_1,q_2)$ line is steeper than the $(q_2,q_3)$ line, so $|q_1{q}_2|>|q_2{q}_3|$, which gives $|q_1|>|q_3|$.

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2. $q_2$ is the least in magnitude, so $q_2<q_3<q_1$. Option (A).

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Answer: $q_2<q_3<q_1$, option (A)

1. $R=\frac{\rho l}{A}$. For Q, doubling the diameter makes the area $4A$, and the length is $l/2$.

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2. $R_Q=\frac{\rho (l/2)}{4A}=\frac{R}{8}$. Option (C).

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Answer: R/8, option (C)

1. $\vec{F}=I\vec{l}\times \vec{B}=0.5(0.01\hat{i})\times (0.4\hat{j}+0.6\hat{k})\times 10^{-3}$ N.

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2. Using $\hat{i}\times \hat{j}=\hat{k}$ and $\hat{i}\times \hat{k}=-\hat{j}$: $\vec{F}=5\times 10^{-6}(0.4\hat{k}-0.6\hat{j})$ N $=(-3\hat{j}+2\hat{k})$ μN. Option (B).

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Answer: $(-3\hat{j}+2\hat{k})$ μN, option (B)

1. Induced charge $q=\frac{\Delta \Phi }{R}=\frac{NBA}{R}$.

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2. $q=\frac{100\times 0.09\times 0.05}{1.5}=0.30$ C. Option (B).

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Answer: 0.30 C, option (B)

1. For a solenoid, $L=\frac{{\mu }_0{N}^2\pi r^2}{l}$, so $N^2=\frac{Ll}{{\mu }_0\pi r^2}$.

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2. $N^2=\frac{0.016\times 0.81}{4\pi \times 10^{-7}\times \pi \times 4\times 10^{-4}}\approx 8.2\times 10^6$, so $N\approx 2866$. Option (B).

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Answer: 2866, option (B)

1. $P_{avg}=V_{rms}{I}_{rms}\cos \varphi =\frac{v_0}{\sqrt{2}}\frac{i_0}{\sqrt{2}}\cos \varphi =\frac{v_0{i}_0}{2}\cos \varphi$. Option (D).

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Answer: $\frac{i_0{v}_0}{2}\cos \varphi$, option (D)

1. $E=hc/\lambda$. P has the shortest wavelength (X-rays), Q the longest (microwaves) and R lies between them (UV). Option (A).

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Answer: X-rays, microwaves, UV, option (A)

1. Apparent depth $=\frac{\text{real depth}}{n}=\frac{H}{4/3}=\frac{3H}{4}$. Option (B).

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Answer: 3H/4, option (B)

1. $e{V}_0=h\nu -{\varphi }_0$, so $V_0=\frac{h}{e}\nu -\frac{{\varphi }_0}{e}$ and the slope is $m=h/e$.

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2. Therefore $e=h/m$. Option (C).

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Answer: h/m, option (C)

1. $\lambda =\frac{h}{mv}$. At the same speed, $\lambda \propto 1/m$, and $m_e<m_p<m_d$, so ${\lambda }_e>{\lambda }_p>{\lambda }_d$. Option (B).

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Answer: ${\lambda }_e>{\lambda }_p>{\lambda }_d$, option (B)

1. B.E./A rises steeply for light nuclei, reaches a maximum of about 8.75 MeV near A = 56 (iron), then decreases slowly for heavy nuclei. This matches graph (A).

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Answer: Option (A)

1. In forward bias, the applied voltage opposes the barrier potential. The effective barrier height falls and the depletion layer narrows. Option (C).

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Answer: Option (C)

1. By Lenz's law, the induced current opposes the motion of the magnet. With more turns the induced emf and current are larger, so the opposition is stronger. Both A and R are true, and R explains A. Option (A).

   1 mark

Answer: Option (A)

1. In a radial field the plane of the coil is always parallel to $B$, so the torque is $NIAB$. At equilibrium, $NIAB=k\varphi$, which gives $\varphi \propto I$. Both A and R are true, and R explains A. Option (A).

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Answer: Option (A)

1. A is true: physically joining two slabs leaves surface roughness far larger than the interatomic spacing, so there is no continuous junction.

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2. R is false: a p-type semiconductor has $n_h\gg n_e$, and an n-type has $n_e\gg n_h$, which is the opposite of what R states. Option (C).

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Answer: Option (C)

1. The potential energy $U=-\frac{k{e}^2}{r}$ is negative, so A is false. The total energy of a bound electron is negative, so R is false. Option (D).

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Answer: Option (D)

1. Terminal voltage of a cell delivering current $I$: $V=E-Ir\Rightarrow E=V+Ir$.

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2. Case 1 ($I=2$ A, $V=5$ V): $E=5+2r$ ...(1). Case 2 ($I=4$ A, $V=4$ V): $E=4+4r$ ...(2).

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3. Equating (1) and (2): $5+2r=4+4r\Rightarrow 2r=1\Rightarrow r=0.5\Omega$.

   ½ mark

4. From (1): $E=5+2(0.5)=6$ V.

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Answer: E = 6 V, r = 0.5 Ω

1. Condition for the $n^{th}$ minimum in single-slit diffraction: $a\sin \theta =n\lambda$. For the first minimum $n=1$: $a\sin \theta =\lambda$.

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2. $a\times \sin 30^{\circ }=600\times 10^{-9}$ m, so $a\times \frac{1}{2}=600\times 10^{-9}$ m.

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3. $a=1200\times 10^{-9}$ m $=1.2\times 10^{-6}$ m $(1.2\mu$m$)$.

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Answer: a = 1.2 × 10⁻⁶ m (1.2 μm)

1. Phase difference $\varphi =\frac{2\pi }{\lambda }\times \Delta x=\frac{2\pi }{\lambda }\times \frac{\lambda }{8}=\frac{\pi }{4}$.

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2. Resultant intensity: $I=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$, with $I_1=I_2=I_o$.

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3. $I=I_o+I_o+2I_o\cos \frac{\pi }{4}=2I_o+2I_o\times \frac{1}{\sqrt{2}}$.

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4. $I=(2+\sqrt{2})I_o\approx 3.414I_o$. (Equivalently, $I=4I_o{\cos }^2\frac{\pi }{8}$.)

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Answer: I = (2 + √2) I₀ ≈ 3.41 I₀

1. The ray travels along the curved surface, so it meets that surface at exactly the critical angle $C$: $\sin C=\frac{1}{n}=\frac{\sqrt{3}}{2}$, so $C=60^{\circ }$.

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2. The end face is perpendicular to the curved surface, so the angle of refraction at the face is $r=90^{\circ }-C=30^{\circ }$.

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3. Snell's law at the air–rod face: $1\times \sin \theta =n\sin r=\frac{2}{\sqrt{3}}\times \sin 30^{\circ }$.

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4. $\sin \theta =\frac{2}{\sqrt{3}}\times \frac{1}{2}=\frac{1}{\sqrt{3}}$, so $\theta ={\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)\approx 35.3^{\circ }$.

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Answer: θ = sin⁻¹(1/√3) ≈ 35.3°

1. Time period of revolution in the $n^{th}$ orbit: $T=\frac{2\pi r_n}{v_n}$ ...(1).

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2. Bohr's quantisation condition: $m{v}_n{r}_n=\frac{nh}{2\pi }\Rightarrow v_n=\frac{nh}{2\pi m{r}_n}$ ...(2). Substituting in (1): $T=\frac{4{\pi }^{2}m{r}_n^2}{nh}$.

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3. Radius of the $n^{th}$ orbit: $r_n=\frac{{\epsilon }_0{n}^2{h}^2}{\pi m{e}^2}$, so $r_n\propto n^2$.

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4. Hence $T=\frac{4{\pi }^{2}m}{nh}\cdot \frac{{\epsilon }_0^2{n}^4{h}^4}{{\pi }^2{m}^2{e}^4}=\frac{4{\epsilon }_0^2{n}^3{h}^3}{m{e}^4}$, so $T\propto n^3$. (Shortcut: $r\propto n^2$ and $v\propto \frac{1}{n}$ give $T=\frac{2\pi r}{v}\propto n^3$.)

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Answer: T = 4ε₀²n³h³/(me⁴), so T ∝ n³

1. Each trivalent dopant atom creates one hole. Holes per m³ $=\frac{5\times 10^{28}}{5\times 10^7}=10^{21}{\text{m}}^{-3}$. Since $1{\text{m}}^3=10^6{\text{cm}}^3$, holes per cm³ $=\frac{10^{21}}{10^6}=10^{15}{\text{cm}}^{-3}$.

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2. Example of such a (trivalent/acceptor) dopant: aluminium, boron, gallium or indium.

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Answer: 10¹⁵ holes per cm³; dopant e.g. aluminium (or B, Ga, In)

1. (i) For two cells in parallel: $E_{eq}=\frac{E_1{r}_2+E_2{r}_1}{r_1+r_2}$.

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2. $E_{eq}=\frac{3\times 0.4+6\times 0.2}{0.2+0.4}=\frac{2.4}{0.6}=4$ V.

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3. (ii) $r_{eq}=\frac{r_1{r}_2}{r_1+r_2}$.

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4. $r_{eq}=\frac{0.2\times 0.4}{0.6}=\frac{0.08}{0.6}\approx 0.133\Omega$.

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5. (iii) $I=\frac{E_{eq}}{R+r_{eq}}=\frac{4}{4+0.133}$.

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6. $I=\frac{4}{4.133}\approx 0.97$ A.

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Answer: (i) 4 V; (ii) ≈ 0.133 Ω (2/15 Ω); (iii) ≈ 0.97 A

1. (i)(i) Stretching does not change the volume: $Al=A^{\prime }(2l)\Rightarrow A^{\prime }=\frac{A}{2}$.

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2. $R^{\prime }=\frac{\rho (2l)}{A/2}=4\frac{\rho l}{A}=4R$.

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3. (i)(ii) Drift velocity: $v_d=\frac{eE\tau }{m}=\frac{eV\tau }{ml}$, since the field is $E=V/l$.

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4. The cell's emf V stays the same while the length doubles, so $v_d^{\prime }=\frac{eV\tau }{m(2l)}=\frac{v_d}{2}$.

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5. (ii) No. Free electrons keep moving randomly in all directions. The field only superimposes a small average drift towards higher potential, so they do not all move in the same direction.

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Answer: (i) R′ = 4R; v_d′ = v_d/2. (ii) No; electrons move randomly with only a small net drift.

1. Take a loop of radius R carrying current I, with P on its axis at distance x from the centre. By Biot–Savart law, an element $d\vec{l}$ gives $d\vec{B}=\frac{{\mu }_{0}I}{4\pi }\frac{d\vec{l}\times \vec{r}}{r^3}$, where $r^2=x^2+R^2$.

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2. $d\vec{l}\perp \vec{r}$ for every element, so $|d\vec{l}\times \vec{r}|=rdl$ and $dB=\frac{{\mu }_{0}Idl}{4\pi (x^2+R^2)}$.

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3. Each $d\vec{B}$ has a component along the axis and one perpendicular to it. The perpendicular components from diametrically opposite elements cancel, so only the axial ($x$) components add.

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4. $d{B}_x=dB\cos \theta$ with $\cos \theta =\frac{R}{\sqrt{x^2+R^2}}$, so $d{B}_x=\frac{{\mu }_{0}IRdl}{4\pi (x^2+R^2{)}^{3/2}}$.

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5. Summing over the loop, $\oint dl=2\pi R$, so $\vec{B}=\frac{{\mu }_{0}I{R}^2}{2(x^2+R^2{)}^{3/2}}\hat{i}$, directed along the axis.

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6. At the centre, $x=0$: $\vec{B}=\frac{{\mu }_{0}I}{2R}\hat{i}$.

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Answer: B on axis = μ₀IR²/[2(x²+R²)^{3/2}] along the axis; at the centre B = μ₀I/(2R)

1. (a) While the current grows, a back emf $\epsilon =-L\frac{dI}{dt}$ opposes it. The source must do work against it at the rate $\frac{dW}{dt}=|\epsilon |I=LI\frac{dI}{dt}$.

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2. So $dW=LIdI$.

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3. Total work to raise the current from 0 to I: $W={\int }_0^{I}LIdI=\frac{1}{2}L{I}^2$. This work is stored as magnetic energy.

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4. (b) For a long air-filled solenoid of length $l$, cross-section A and n turns per unit length: $L={\mu }_0{n}^{2}Al$ and $B={\mu }_{0}nI$, so $I=\frac{B}{{\mu }_{0}n}$.

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5. $U_B=\frac{1}{2}L{I}^2=\frac{1}{2}{\mu }_0{n}^{2}Al{\left(\frac{B}{{\mu }_{0}n}\right)}^2=\frac{B^{2}Al}{2{\mu }_0}$.

   ½ mark

6. Energy density $u_B=\frac{U_B}{\text{volume}}=\frac{U_B}{Al}=\frac{B^2}{2{\mu }_0}$.

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Answer: (a) W = ½LI²; (b) u_B = B²/(2μ₀)

1. (a) The total current is $I=I_c+I_d$. In the connecting wires (outside the capacitor) there is no changing electric field, so $I_d=0$ and $I=I_c$.

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2. Between the plates no charge crosses the gap, so $I_c=0$ and $I=I_d={\epsilon }_0\frac{d{\Phi }_E}{dt}={\epsilon }_0\frac{d(EA)}{dt}$.

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3. The field between the plates is $E=\frac{Q}{{\epsilon }_{0}A}$, so $I_d={\epsilon }_{0}A\frac{d}{dt}\left(\frac{Q}{{\epsilon }_{0}A}\right)=\frac{dQ}{dt}$.

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4. $\frac{dQ}{dt}$ is exactly the conduction current $I_c$ in the wires. So $I_d$ (between the plates) equals $I_c$ (in the wires), and $I_c+I_d$ has the same value at every point of the circuit.

   ½ mark

5. (b) Yes. At each plate the conduction current $I_c$ entering the plate equals the displacement current $I_d$ leaving it through the gap. With $I_d$ included, current into the junction equals current out of it, so the junction rule holds at each plate.

   1 mark

Answer: (a) I_d = dQ/dt = I_c, so I_c + I_d is the same everywhere; (b) Yes, because I_c into a plate equals I_d leaving it.

1. (a) Each photon has the same energy $h\nu$, but electrons are bound in the metal with different energies. Electrons at the surface need only the work function to escape, while deeper or more tightly bound electrons lose more energy. So the emitted electrons have kinetic energies from zero up to a maximum $K_{max}=h\nu -{\varphi }_0$.

   1 mark

2. (b) Saturation current means all emitted electrons are collected, so it measures the number of photoelectrons emitted per second. This number is directly proportional to the number of photons falling per second, i.e. to the intensity. Different intensities therefore give different saturation currents.

   1 mark

3. (c) As λ increases, the frequency $\nu =c/\lambda$ and the photon energy $h\nu$ decrease. When λ exceeds the threshold wavelength ${\lambda }_0$ (i.e. $\nu <{\nu }_0$), $h\nu$ is less than the work function. No electron can then escape, whatever the intensity, so emission stops.

   1 mark

Answer: (a) electrons have different binding energies, so KE ranges from 0 to K_max; (b) saturation current ∝ photoelectrons per second ∝ intensity; (c) beyond the threshold wavelength λ₀, hν < work function, so no emission.

1. (a) Mass defect: the difference between the total mass of the separate nucleons and the actual mass of the nucleus, $\Delta M=[Z{m}_p+(A-Z)m_n]-M$.

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2. Binding energy: the minimum energy needed to separate a nucleus completely into its constituent nucleons, $E_b=\Delta M{c}^2$.

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3. Fission: a heavy nucleus (A ≈ 240, $E_{bn}$ ≈ 7.6 MeV) splits into two medium nuclei (A ≈ 120, $E_{bn}$ ≈ 8.5 MeV). The binding energy per nucleon increases, so the nucleons become more tightly bound and the difference (≈ 200 MeV per fission) is released.

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4. (b) $\Delta m=(m_p+m_n)-m_d=(1.007277+1.008665)-2.013553$ u $=2.015942-2.013553=0.002389$ u.

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5. Energy equivalent $=\Delta m\times 931.5$ MeV/u $=0.002389\times 931.5\approx 2.225$ MeV ($\approx 2.22$ MeV).

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Answer: Δm = 0.002389 u; energy ≈ 2.225 MeV (≈ 2.22 MeV)

1. (a) Forward bias circuit: a battery with a rheostat (potential divider) is connected across the diode, with the p-side to the + terminal. A milliammeter is in series with the diode and a voltmeter is across it. By varying the rheostat, readings of V and I are taken.

   ½ mark

2. Reverse bias circuit: same arrangement with a higher-voltage battery and the n-side to the + terminal (p-side to −), using a microammeter in series and a voltmeter across the diode.

   ½ mark

3. (b) The graph has I (mA) on the +y axis against forward V on the +x axis, and I (μA) on the −y axis against reverse V on the −x axis. Forward: current is almost zero until the threshold (knee) voltage (≈ 0.7 V for Si, 0.3 V for Ge), then rises steeply, almost exponentially. Reverse: a very small, nearly constant current (μA) flows until the breakdown voltage, where the current suddenly rises sharply.

   1 mark

4. (c) Any two: (1) the threshold/knee voltage of the diode; (2) the reverse saturation current; (3) the breakdown voltage; (4) the diode has very low resistance in forward bias and very high resistance in reverse bias (it conducts in one direction only).

   1 mark

Answer: Forward and reverse bias circuits described; characteristic with knee at ≈0.7 V (Si) and reverse breakdown; information: knee voltage, reverse saturation current, breakdown voltage, low forward / high reverse resistance.

1. (i) Final charge $Q=CV=500\times 10^{-6}\times 10=5\times 10^{-3}$ C $=5$ mC. Option (B).

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2. (ii) With $S_1$ open and $S_2$ closed, the battery is removed and the capacitor discharges completely through R. Final charge = zero. Option (A).

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3. (iii) RC is the time constant, so it has dimensions of time: $[M^0{L}^{0}T{A}^0]$. Option (D).

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4. (iv)(a) $\tau =RC=20\times 10^3\times 500\times 10^{-6}=10$ s. $I=\frac{dq}{dt}=\frac{V}{R}{e}^{-t/RC}=\frac{10}{20\times 10^3}{e}^{-5/10}=0.5e^{-1/2}$ mA $=\frac{1}{2\sqrt{e}}$ mA. Option (A).

   1 mark

5. (iv)(b) [alternative] At $t=0$: $I_0=\frac{V}{R}=\frac{10}{20\times 10^3}=0.5$ mA. Option (B).

Answer: (i) B, 5 mC; (ii) A, zero; (iii) D, [M⁰L⁰TA⁰]; (iv)(a) A, 1/(2√e) mA; (iv)(b) B, 0.5 mA

1. (i) A convex lens forms a real image that is inverted both up–down and left–right. So the arrow points down and the ball appears on the opposite side (left). Option (C).

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2. (ii) A convex mirror always forms virtual, erect images, so its magnification is always positive. Statement (A) is therefore incorrect. Option (A).

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3. (iii)(a) Cutting perpendicular to the principal axis gives two plano-convex halves, each with one curved surface: $\frac{1}{f^{\prime }}=(n-1)\frac{1}{R}=\frac{1}{2}\cdot \frac{1}{f}$, so $f^{\prime }=2f$. Option (B).

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4. (iii)(b) [alternative] $u=-20$ cm, $v=50-20=+30$ cm. $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{30}+\frac{1}{20}=\frac{5}{60}$, so $f=12$ cm. Option (B).

5. (iv) From Newton's formula, with object distance $f+X_1$ and image distance $f+X_2$: $\frac{1}{f+X_2}+\frac{1}{f+X_1}=\frac{1}{f}$ simplifies to $X_1{X}_2=f^2$, so $f=\sqrt{X_1{X}_2}$. Option (C).

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Answer: (i) C; (ii) A; (iii)(a) B, 2f; (iii)(b) B, 12 cm; (iv) C, √(X₁X₂)

1. (i) The potential of the external field: $V(r)=-{\int }_{\infty }^r\vec{E}\cdot d\vec{r}=\frac{A}{r}=\frac{3\times 10^5}{r}$.

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2. Without the field, $U_i=\frac{k{q}_1{q}_2}{r_{12}}$. With the field, $U_f=\frac{k{q}_1{q}_2}{r_{12}}+q_{1}V(r_1)+q_{2}V(r_2)$.

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3. So $\Delta U=q_{1}V(r_1)+q_{2}V(r_2)$, with $r_1=r_2=3$ cm $=3\times 10^{-2}$ m, giving $V=\frac{3\times 10^5}{3\times 10^{-2}}=10^7$ V at both charges.

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4. $\Delta U=(5\times 10^{-6}-1\times 10^{-6})\times 10^7=40$ J.

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5. (ii)(1) $C=\frac{Q}{V}=\frac{80\mu \text{C}}{16\text{V}}=5\mu$F.

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6. (ii)(2) With the dielectric, $C^{\prime }=KC=3\times 5=15\mu$F. The charge stays 80 μC, so $V^{\prime }=\frac{Q}{C^{\prime }}=\frac{80}{15}\approx 5.33$ V.

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7. (ii)(3) No. Capacitance depends only on the size, shape and separation of the conductors and on the medium. Doubling the charge doubles V (to 32 V), so $C=Q/V$ stays 5 μF.

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Answer: (i) ΔU = 40 J; (ii)(1) 5 μF; (2) ≈ 5.33 V; (3) No, capacitance depends on geometry and medium, not charge

1. (i) At $r=3R$ (outside each shell) the field is that of the total charge enclosed, as if placed at the centre: $E=\frac{k{Q}_{total}}{r^2}$.

   ½ mark

2. Total charges: A: $6q-2q=4q$; B: $-4q+8q=4q$; C: $14q-10q=4q$.

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3. $E=\frac{k(4q)}{(3R{)}^2}=\frac{4kq}{9R^2}$ for each, so $E_A:E_B:E_C=1:1:1$ (all equal).

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4. (ii) From the figure, $DC=5$ cm, $BC=BA=5$ cm and $DA=15$ cm, with +6 μC at D and −6 μC at B. $V_C=\frac{k(6\times 10^{-6})}{0.05}-\frac{k(6\times 10^{-6})}{0.05}=0$.

   1 mark

5. $V_A=k(6\times 10^{-6})\left(\frac{1}{0.15}-\frac{1}{0.05}\right)=9\times 10^9\times 6\times 10^{-6}\times \left(-\frac{40}{3}\right)$.

   ½ mark

6. $V_A=-7.2\times 10^5$ V.

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7. Work done (by external agent) on the +5 μC charge: $W=q(V_A-V_C)=5\times 10^{-6}\times (-7.2\times 10^5-0)=-3.6$ J. The work done by the electric field is +3.6 J. The work is path independent, so moving along the semicircle makes no difference.

   1 mark

Answer: (i) E_A = E_B = E_C = 4kq/(9R²), ratio 1:1:1; (ii) W = −3.6 J

1. (i) The force on the proton is $\vec{F}=q(\vec{v}\times \vec{B})$ and points towards the centre of curvature of the path. Near P the path bends upward (force upward) while v is to the right, so B is into the plane of the paper.

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2. Near Q the path still bends upward (force towards the upper side), so B is again into the plane of the paper.

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3. Near R the path bends the other way (force downward), so B is out of the plane of the paper.

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4. The magnetic force does no work, so the speed is constant. Radius of curvature $r=\frac{mv}{qB}$, so $B\propto \frac{1}{r}$.

   ½ mark

5. Near Q the path is least curved (largest radius), so B is smallest there. Near P the curvature is moderate. Near R the path is most sharply curved (smallest radius), so B is largest.

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6. Hence $B_Q<B_P<B_R$.

   ½ mark

7. (ii) The field at the centre of the loop is $B=\frac{{\mu }_{0}I}{2r}$, so $I=\frac{2Br}{{\mu }_0}$. Area $A=\pi r^2$, so $r=\sqrt{\frac{A}{\pi }}$.

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8. Magnetic moment $m=IA=\frac{2Br}{{\mu }_0}A=\frac{2BA}{{\mu }_0}\sqrt{\frac{A}{\pi }}$. Hence shown.

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Answer: (i) B into the page near P and Q, out of the page near R; B_Q < B_P < B_R. (ii) m = (2BA/μ₀)√(A/π)

1. (i) Set-up: Take a rectangular loop PQRS with sides $PQ=RS=b$ and $QR=SP=a$. It carries current $I$ in a uniform field $\vec{B}$. Let $\hat{n}$ be the normal to the loop, making an angle $\theta$ with $\vec{B}$. The sides $b$ are perpendicular to $\vec{B}$. *(Diagram: loop with $\vec{B}$ in its plane direction, normal $\hat{n}$ at angle $\theta$, forces ${\vec{F}}_1$ and ${\vec{F}}_2$ on the arms of length $b$ in opposite directions.)*

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2. The forces on the arms of length $a$ are equal, opposite and along the same line (the axis of the loop). They cancel and produce no torque.

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3. The forces on the arms of length $b$ are equal and opposite: $F_1=F_2=IbB$ [since $\vec{F}=I\vec{l}\times \vec{B}$ and $b\perp \vec{B}$]. Their lines of action are different, so they form a couple.

   ½ mark

4. Each force acts at a distance $\frac{a}{2}$ from the axis, and its lever arm is $\frac{a}{2}\sin \theta$. So
   $\tau =F_1\frac{a}{2}\sin \theta +F_2\frac{a}{2}\sin \theta$

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5. $\tau =IbBa\sin \theta =I(ab)B\sin \theta =IAB\sin \theta$, where $A=ab$ is the area of the loop.

   ½ mark

6. With $\vec{m}=IA\hat{n}$, in vector form: $\vec{\tau }=I\vec{A}\times \vec{B}=\vec{m}\times \vec{B}$. For a coil of $N$ turns, $\tau =NIAB\sin \theta$.

   ½ mark

7. (ii)(1) Radius: The magnetic force provides the centripetal force: $qvB=\frac{m{v}^2}{r}$, so $r=\frac{mv}{qB}=\frac{\sqrt{2mK}}{qB}$. Hence $r\propto \sqrt{K}$.

   ½ mark

8. When the kinetic energy becomes $\frac{K}{2}$:
   $\frac{r^{\prime }}{r}=\sqrt{\frac{K/2}{K}}=\frac{1}{\sqrt{2}}$, so $r^{\prime }=\frac{r}{\sqrt{2}}$. The radius becomes $\frac{1}{\sqrt{2}}$ times (about 0.707 times) the original.

   ½ mark

9. (ii)(2) Time period: $T=\frac{2\pi r}{v}=\frac{2\pi m}{qB}$.

   ½ mark

10. $T$ does not depend on the speed or the kinetic energy of the particle. So the time period of revolution remains unchanged.

    ½ mark

Answer: (i) $\vec{\tau }=I\vec{A}\times \vec{B}=\vec{m}\times \vec{B}$, magnitude $IAB\sin \theta$; (ii)(1) radius becomes $r/\sqrt{2}$; (2) time period stays the same.