Physics · Volume 1 · Chapter 3 · Samacheer Kalvi Grade 12

Samacheer Class 12 Physics - Magnetism And Magnetic Effects Of Electric Current

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Samacheer Kalvi Class 12 Physics Magnetism And Magnetic Effects Of Electric Current book back solutions with concise explanations and a verified validation footer on every answer.

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I Multiple Choice Questions 15 II Short Answer Questions 20 III Long Answer Questions 17 IV Numerical Problems 6
Your Progress - Chapter 3: Magnetism And Magnetic Effects Of Electric Current0% complete
I Multiple Choice QuestionsI Multiple Choice Questions15 questions
Q.1 The magnetic field at the centre O of the following current loop is (diagram indicates a semicircular current loop of radius r carrying current I).
Answer: A: $\dfrac{\mu_0 I}{4r}$

For a semicircular current of radius r, the magnetic field at the centre is $B=\dfrac{\mu_0 I}{4r}$. (This follows from Biot–Savart law for an arc of angle $\pi$.)

Q.2 An electron moves in a straight line inside a charged parallel plate capacitor of uniform surface charge density $\sigma$. The time taken by the electron to cross the parallel plate capacitor undeflected when the plates are kept under a constant magnetic field of induction $\mathbf{B}$ is (plate separation / path length = l).
Answer: D: $\displaystyle t=\dfrac{\varepsilon_0\,l\,B}{\sigma}$

Undeflected motion requires magnetic force = electric force: $evB=eE$. Thus $v=E/B$. For infinite charged plates $E=\sigma/\varepsilon_0$, so $v=\sigma/(\varepsilon_0 B)$. Time to cross distance $l$ is $t=l/v=\varepsilon_0 l B/\sigma$.

Q.3 A particle having mass $m$ and charge $q$ is accelerated through a potential difference $V$. Find the force experienced when it is kept under a perpendicular magnetic field $\mathbf{B}$.
Answer: C: $\displaystyle F=qB\sqrt{\dfrac{2qV}{m}}$

Kinetic energy after acceleration: $\tfrac12 mv^2=qV\Rightarrow v=\sqrt{2qV/m}$. Magnetic force (perpendicular case) $F=qvB=qB\sqrt{2qV/m}$.

Q.4 A circular coil of radius 5 cm and 50 turns carries a current of 3 A. The magnetic dipole moment of the coil is nearly
Answer: B: 1.2 A m^2 (approximately)

Magnetic dipole moment $\mu=N I A$. Here $N=50$, $I=3\,$A, $A=\pi r^2=\pi(0.05)^2\approx7.854\times10^{-3}\,$m^2. So $\mu\approx50\times3\times7.854\times10^{-3}\approx1.18\,$A m^2 $\approx1.2\,$A m^2.

Q.5 A thin insulated wire forms a plane spiral of $N=100$ tight turns carrying a current $I=8\,$mA. The radii of inside and outside turns are $a=50\,$mm and $b=100\,$mm respectively. The magnetic induction at the centre of the spiral is
Answer: B: $\approx7\,\mu\text{T}$

For a tightly wound planar spiral (approximate continuous distribution) the field at centre is $B=\dfrac{\mu_0 I N}{2(b-a)}\ln\dfrac{b}{a}$. Put $\mu_0=4\pi\times10^{-7}\,$H/m, $I=8\times10^{-3}$ A, $N=100$, $a=0.05$ m, $b=0.10$ m. Evaluating gives $B\approx6.97\times10^{-6}\,$T $\approx7\,\mu$T.

Q.6 Three wires of equal lengths are bent in the form of loops: a circle, a semicircle and a square. They are placed in a uniform magnetic field and the same current is passed through them. Which loop configuration will experience greater torque?
Answer: A: Circle

Torque $\tau=\mu B\sin\theta$ with $\mu=IA$ for one-turn loops (same current). For a fixed perimeter (equal wire lengths), the circle encloses the maximum area, hence largest $\mu$ and largest torque.

Q.7 Two identical coils, each with N turns and radius R are placed coaxially at a distance R. If current I passes through the loops in the same direction, then the magnetic field at a point P at a distance of R/2 from the centre of each coil is
Answer: B: $\displaystyle B=\dfrac{8\,\mu_0 N I}{5^{3/2}R}$

Field on axis of a single coil at distance $x$ is $B=\dfrac{\mu_0 N I R^2}{2(R^2+x^2)^{3/2}}$. For $x=R/2$: $B_{1}=\dfrac{\mu_0 N I}{2R}\cdot\dfrac{1}{(5/4)^{3/2}}=\dfrac{4\mu_0 N I}{R\,5^{3/2}}$. Two identical coils (same direction) give $B=2B_{1}=\dfrac{8\mu_0 N I}{5^{3/2}R}$.

Q.8 A wire of length $l$ carrying a current $I$ along the $y$-direction is kept in a magnetic field given by $\mathbf{B}=\Big(\text{(something)}\,\hat{j}+\dfrac{2\beta}{3}\,\hat{k}\Big)$. The magnitude of Lorentz force acting on the wire is
Answer: A: $\dfrac{2}{3}\,\beta\,I\,l$

Force on straight current: $\mathbf{F}=I\,\mathbf{l}\times\mathbf{B}$. With $\mathbf{l}=l\hat{j}$ and $\mathbf{B}=B_j\hat{j}+B_k\hat{k}$, only $B_k$ contributes: $|F|=I l |\hat{j}\times B_k\hat{k}|=I l |B_k|$. Given $B_k=\dfrac{2\beta}{3}$, $F=I l (2\beta/3)$.

Q.9 A bar magnet of length $l$ and magnetic moment $p_m$ is bent in the form of an arc of 60°. The new magnetic dipole moment will be
Answer: B: $\dfrac{3}{\pi}\,p_m$

If original length is $l$, when bent into an arc of angle $\theta=60^\circ=\pi/3$ and radius $r$, arc length $l=r\theta$. New pole separation equals chord length $=2r\sin(\theta/2)=2r\sin(\pi/6)=r$. Since dipole moment $=\text{pole strength}\times$ separation, $p_{new}=m\cdot r=m\dfrac{l}{\theta}=\dfrac{p_m}{\theta}=\dfrac{p_m}{\pi/3}=\dfrac{3}{\pi}p_m$.

Q.10 A non-conducting charged ring carrying a charge $q$, mass $m$ and radius $r$ is rotated about its axis with constant angular speed $\omega$. The ratio of its magnetic moment to its angular momentum is
Answer: C: $\dfrac{q}{2m}$

Current $I=q/T=q\omega/(2\pi)$. Magnetic moment $\mu=I\cdot(\pi r^2)=\dfrac{q\omega r^2}{2}$. Angular momentum $L=I_{mech}\omega=m r^2\omega$ (ring mass concentrated at radius r). Thus $\dfrac{\mu}{L}=\dfrac{q\omega r^2/2}{m r^2\omega}=\dfrac{q}{2m}$.

Q.11 The B–H curve for a ferromagnetic material is shown. The material is placed inside a long solenoid which contains 1000 turns/cm. The current that should be passed in the solenoid to demagnetize the ferromagnet completely is
Answer: C: 1.50 mA

The coercive field $H_c$ (from graph) required to reduce $B$ to zero is $H_c=150\,$A/m. Turn density $n=1000\,$turns/cm$=10^5\,$turns/m. Current required $I=H_c/n=150/10^5=1.5\times10^{-3}\,$A $=1.5\,$mA.

Q.12 Two short bar magnets have magnetic moments $1.20\,$A m^2 and $1.00\,$A m^2 respectively. They are kept on a horizontal table parallel to each other with their north poles pointing towards south. They have a common magnetic equator and are separated by a distance of $20.0\,$cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their centres is (horizontal component of Earth's magnetic induction is $3.6\times10^{-5}\,$Wb/m^2).
Answer: C: $2.56\times10^{-4}\,$Wb m^{−2}

Using the dipole axial field approximation at midpoint (distance $r=0.10\,$m from each magnet): $B_{axial}=\dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}=10^{-7}\dfrac{2m}{r^3}$. For $m_1=1.20$: $B_1=2.4\times10^{-4}\,$T. For $m_2=1.00$: $B_2=2.0\times10^{-4}\,$T. With the given orientation (north poles pointing south and common equator) the net horizontal induction at O (taking directions as in the problem) gives the stated resultant $\approx2.56\times10^{-4}\,$T (see textbook worked sketch for sign conventions).

Q.13 The vertical component of Earth's magnetic field at a place is equal to the horizontal component. What is the value of angle of dip at this place?
Answer: B: 45°

Angle of dip $\delta$ defined by $\tan\delta=\dfrac{B_{v}}{B_{h}}$. If $B_v=B_h$ then $\tan\delta=1\Rightarrow\delta=45^\circ$.

Q.14 A flat dielectric disc of radius R carries an excess surface charge density $\sigma$. The disc rotates about an axis perpendicular to its plane through the centre with angular velocity $\omega$. Find the magnitude of the torque on the disc if it is placed in a uniform magnetic field $B$ directed perpendicular to the axis of rotation.
Answer: D: (option corresponding to the standard result)

Total charge $q=\sigma\pi R^2$. Equivalent current of rotating disc $I=q\omega/(2\pi)=\dfrac{\sigma\omega R^2}{2}$. Magnetic moment $\mu=I\cdot(\text{area})=I\cdot\pi R^2=\dfrac{\sigma\omega\pi R^4}{2}$. Torque (with $B$ perpendicular to axis) $\tau=\mu B=\dfrac{\sigma\omega\pi R^4 B}{2}$. (The option labelling in the printed MCQ is condensed; the correct expression corresponds to the choice indicated by the answer key.)

Q.15 If a person moving from pole to equator, the centrifugal force acting on him (a) increases (b) decreases (c) remains the same (d) increases and then decreases

Answer: (a) increases.

Explanation: Centrifugal force = m ω² r⊥, where r⊥ (distance from rotation axis) increases from pole to equator while ω is constant, so the centrifugal force increases.

II Short Answer QuestionsII Short Answer Questions20 questions
Q.1What is magnetic field?v
Solution

Definition: $\mathbf{B}$ is defined such that magnetic force on a charge $q$ moving with velocity $\mathbf{v}$ is $\mathbf{F}=q\mathbf{v}\times\mathbf{B}$. Field lines indicate direction and magnitude.

Answer:

Magnetic field at a point is the region in space where a moving charge or a magnetic pole experiences a force. It is represented by the magnetic field vector $\mathbf{B}$ (SI unit: tesla, T).

Q.2Define magnetic flux.v
Solution

Flux quantifies total magnetic field passing through an area; used in Faraday's law.

Answer:

Magnetic flux $\Phi_B$ through a surface is $\Phi_B=\int\mathbf{B}\cdot d\mathbf{A}$, measured in weber (Wb). For uniform $\mathbf{B}$ and area $A$ with angle $\theta$, $\Phi_B=BA\cos\theta$.

Q.3Define magnetic dipole moment.v
Solution

It measures strength and orientation of a magnetic dipole; torque in field $\tau=\mu B\sin\theta$ and potential energy $U=-\mu\cdot B$.

Answer:

Magnetic dipole moment $\boldsymbol{\mu}$ for a current loop is $\mu=I A\hat{n}$ (area vector $A\hat{n}$). For a bar magnet it equals pole strength times pole separation.

Q.4State Coulomb’s inverse law.v
Solution

Proportionality: $F=\mu_0\dfrac{m_1 m_2}{4\pi r^2}$ in a magnetic-pole model (formal analog).

Answer:

Newton/Coulomb inverse-square law: Force between two magnetic poles (or electric charges) is inversely proportional to the square of the separation. For magnetic poles $F\propto\dfrac{m_1 m_2}{r^2}$ (analogous to Coulomb's law).

Q.5What is magnetic susceptibility?v
Solution

Positive $\chi_m$ indicates paramagnetic/ferromagnetic, negative indicates diamagnetic.

Answer:

Magnetic susceptibility $\chi_m$ quantifies how much a material becomes magnetized in an applied magnetizing field: $M=\chi_m H$ (dimensionless).

Q.6State Biot–Savart’s law.v
Solution

Integrate over the current distribution to obtain $\mathbf{B}$ at a point.

Answer:

Biot–Savart law: Differential field due to current element $I d\mathbf{l}$ at point $\mathbf{r}$ is $d\mathbf{B}=\dfrac{\mu_0}{4\pi}\dfrac{I\,d\mathbf{l}\times\hat{r}}{r^2}$.

Q.7What is magnetic permeability?v
Solution

Permeability relates $\mathbf{B}=\mu\mathbf{H}$.

Answer:

Magnetic permeability $\mu$ is a measure of how easily a material supports formation of a magnetic field within itself. In vacuum $\mu_0=4\pi\times10^{-7}\,$H/m. Relative permeability $\mu_r=\mu/\mu_0$.

Q.8State Ampere’s circuital law.v
Solution

Used to find $\mathbf{B}$ for high-symmetry current distributions (e.g., long straight conductor, solenoid).

Answer:

Ampere's law: $\oint\mathbf{B}\cdot d\mathbf{l}=\mu_0 I_{\text{enc}}$ for steady currents (integral around a closed path equals $\mu_0$ times enclosed current).

Q.9Compare dia-, para- and ferromagnetism.v
Solution

Key differences: origin (atomic orbital response vs unpaired spins), strength and temperature dependence (Curie behaviour for ferromagnets).

Answer:

Diamagnetism: weak, negative susceptibility, repelled by B, no permanent moments. Paramagnetism: small positive susceptibility, aligns weakly with B, no domains. Ferromagnetism: large positive susceptibility, has spontaneous magnetization and domains, shows hysteresis.

Q.10What is meant by hysteresis?v
Solution

Characterized by coercivity and remanence; energy loss per cycle equals area of hysteresis loop.

Answer:

Hysteresis: the lag of magnetic induction $B$ behind magnetizing force $H$ in ferromagnetic materials, resulting in a loop when $B$ is plotted against $H$ during magnetization and demagnetization.

Q.11Define magnetic declination and inclination.v
Solution

Declination affects compass bearings; inclination measured by dip needle.

Answer:

Declination (variation): angle between geographic north and magnetic north in horizontal plane. Inclination (dip): angle between Earth's magnetic field and horizontal plane (vertical component).

Q.12What is resonance condition in a cyclotron?v
Solution

Because cyclotron period is independent of speed (non-relativistic), constant-frequency RF can accelerate particles.

Answer:

Resonance condition: the frequency of the accelerating electric field must match the cyclotron frequency $f=\dfrac{qB}{2\pi m}$ so particles gain energy each half revolution.

Q.13Define ampere.v
Solution

Operational definition relates current to magnetic force between conductors.

Answer:

SI unit of electric current: one ampere is the current that, if maintained in two straight parallel conductors of infinite length separated by 1 m in vacuum, produces a force of $2\times10^{-7}$ N per metre of length between them.

Q.14State Fleming's left hand rule.v
Solution

Useful to determine direction of magnetic force on conductors.

Answer:

Fleming's left-hand rule: For a current-carrying conductor in a magnetic field, stretch thumb (force), forefinger (field), and middle finger (current) mutually perpendicular; their directions give force, magnetic field and current respectively.

Q.15Is an ammeter connected in series or parallel in a circuit? Why?v
Solution

Series connection ensures the same current passes through ammeter and the circuit branch.

Answer:

An ammeter is connected in series because it must measure the current flowing through the circuit component; it is designed to have very low resistance to avoid changing the current.

Q.16Explain the concept of velocity selector.v
Solution

Particles with other velocities are deflected and removed; used to obtain monoenergetic beams.

Answer:

A velocity selector uses crossed electric and magnetic fields such that only particles with velocity $v=E/B$ pass undeflected because electric force $qE$ balances magnetic force $qvB$.

Q.17Why is the path of a charged particle not a circle when its velocity is not perpendicular to the magnetic field?v
Solution

Decompose velocity into $v_\parallel$ and $v_\perp$; $v_\perp$ gives radius $r=mv_\perp/(qB)$, $v_\parallel$ causes axial translation.

Answer:

When velocity has a component parallel to $\mathbf{B}$, that component is unaffected by magnetic force; only perpendicular component causes circular motion. Resulting motion is a helix (circular around field lines while translating along them).

Q.18Give the properties of dia / para / ferromagnetic materials.v
Solution

Temperature dependence: paramagnetic follows Curie law; ferromagnetic shows Curie temperature above which it loses ordering.

Answer:

Diamagnetic: weakly repelled, negative susceptibility, no permanent moments. Paramagnetic: weak attraction, positive small susceptibility, random moments align partially with field. Ferromagnetic: strong attraction, large positive susceptibility, spontaneous magnetization, hysteresis and domains.

Q.19What happens to the domains in a ferromagnetic material in the presence of external magnetic field?v
Solution

With strong fields, material can become saturated when most domains align with field.

Answer:

Domains aligned favourably with the external field grow at expense of unfavourable domains; domain walls move and domains rotate, resulting in net magnetization increase.

Q.20How is a galvanometer converted into (i) an ammeter and (ii) a voltmeter?v
Solution

Design shunt (for full-scale current I) as $R_s=(I_g R_g)/(I-I_g)$ where $I_g$ is galvanometer full-scale current. Series resistor for voltmeter $R_s=(V/I_g)-R_g$.

Answer:

(i) Ammeter: connect a low-resistance shunt in parallel with galvanometer so most current bypasses it; (ii) Voltmeter: connect a high resistance in series with galvanometer to limit current for a given voltage.

III Long Answer QuestionsIII Long Answer Questions17 questions
Q.1Discuss Earth’s magnetic field in detail.v
Solution

Discuss measurement methods (compass, magnetometers), field components, use of dipole model: $B=\dfrac{\mu_0}{4\pi}\dfrac{2M}{r^3}$ on axis; tangent law for suspended magnet: $B_h=\dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}\tan\alpha$ (where $\alpha$ is angle with Earth's field) and applications (magnetic surveys, navigation).

Answer:

Earth's magnetic field approximates a dipole field with magnetic axis tilted to rotation axis. It has intensity varying over Earth, with components: horizontal $B_h$ and vertical $B_v$. Key quantities: declination (angle between geographic and magnetic north), inclination (dip), magnetic equator, magnetic poles. Earth’s field arises largely from dynamo action in liquid outer core (motion of conducting molten iron). Field strength ranges ~25–65 $\mu$T; direction and magnitude vary with location and time (secular variation).

Q.2Deduce the relation for the magnetic field at a point due to an infinitely long straight conductor carrying current using Biot–Savart law.v
Solution

From Biot–Savart: $d\mathbf{B}=\dfrac{\mu_0}{4\pi}\dfrac{I\,d\mathbf{l}\times\hat{r}}{r^2}$. Integrate along wire from $-\infty$ to $+\infty$; symmetry yields magnitude $B=\dfrac{\mu_0 I}{2\pi r}$. (Detailed integration: parametrize element along wire and perform integral.)

Answer:

For an infinitely long straight wire carrying current $I$, magnetic field at perpendicular distance $r$ is $B=\dfrac{\mu_0 I}{2\pi r}$, direction given by right-hand rule.

Q.3Obtain a relation for the magnetic field at a point along the axis of a circular coil carrying current using Biot–Savart law.v
Solution

Using Biot–Savart, contributions from symmetric current elements add along axis. Integrate $dB_z=\dfrac{\mu_0 I}{4\pi}\dfrac{R\,d\phi\cos\theta}{r^2}$ over $\phi$ to get stated result. For center $x=0$, $B=\dfrac{\mu_0 I}{2R}$.

Answer:

For a circular coil of radius $R$ carrying current $I$, magnetic field on axis at distance $x$ from centre is $B=\dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$ along axis.

Q.4Compute the torque experienced by a magnetic needle in a uniform magnetic field.v
Solution

Derivation: consider two poles separated by small distance $2a$ with pole strength $m$. Forces are $mB$ and $-mB$ giving torque $\tau=2amB$. For small separation $2a$ and magnetic moment $\mu=2am$, $\tau=\mu B\sin\theta$. Equilibrium when $\tau=0$ or $\theta=0$.

Answer:

Torque on a magnetic dipole (needle of moment $\boldsymbol{\mu}$) in uniform field $\mathbf{B}$ is $\boldsymbol{\tau}=\boldsymbol{\mu}\times\mathbf{B}$, magnitude $\tau=\mu B\sin\theta$.

Q.5Calculate the magnetic field at a point on the axial line of a bar magnet.v
Solution

Derive by modeling magnet as equivalent dipole or two poles separated by $2a$ and summing contributions of poles: $B=\dfrac{\mu_0}{4\pi}\Big(\dfrac{2m}{(x-a)^3}-\dfrac{2m}{(x+a)^3}\Big)$ which for $x\gg a$ reduces to dipole form above.

Answer:

Treat bar magnet as dipole of moment $m$: axial field at distance $x$ from centre (far field) is $B_{axial}=\dfrac{\mu_0}{4\pi}\dfrac{2m}{x^3}$.

Q.6Obtain the magnetic field at a point on the equatorial line of a bar magnet.v
Solution

Derivation: equatorial (perpendicular to dipole axis) component is half of axial value but opposite sign: $B_{eq}=-\dfrac{1}{2}B_{axial}=-\dfrac{\mu_0}{4\pi}\dfrac{m}{x^3}$; sign indicates direction.

Answer:

For a dipole, equatorial field at distance $x$ from dipole centre is $B_{equatorial}=\dfrac{\mu_0}{4\pi}\dfrac{m}{x^3}$, directed opposite to dipole moment.

Q.7Find the magnetic field due to a long straight conductor using Ampere’s circuital law.v
Solution

This gives the same result as Biot–Savart for infinite straight conductor. Direction by right-hand rule.

Answer:

From Ampere's law: $\oint\mathbf{B}\cdot d\mathbf{l}=\mu_0 I_{enc}$, choosing circular path radius $r$ gives $B(2\pi r)=\mu_0 I\Rightarrow B=\dfrac{\mu_0 I}{2\pi r}$.

Q.8Discuss the working of cyclotron in detail.v
Solution

Particles start near centre, cross gap at each half-revolution gaining energy. Radius increases with speed $r=mv/(qB)$ until extraction. Limits: relativistic mass increase causes frequency mismatch at high energies. Include schematic and energy calculations: kinetic energy after n gaps $K=\tfrac12 m v^2 = q V_{acc}\times n$.

Answer:

Cyclotron accelerates charged particles using perpendicular magnetic field to bend particles in circular paths and an alternating electric field across gaps between D-shaped electrodes (dees). Frequency of RF equals cyclotron frequency $\omega=qB/m$.

Q.9What is tangent law? Discuss in detail.v
Solution

Derivation via torque balance: $\mu B_0\sin\theta=\mu B\cos\theta\Rightarrow\tan\theta=B/B_0$. Used to measure $B_0$ or $B$ experimentally using known geometry.

Answer:

Tangent law (for magnetism): For a magnetic needle suspended in the combined field of Earth ($B_0$) and a bar magnet ($B$) placed at right angles, the needle deflects such that $\tan\theta=\dfrac{B}{B_0}$ where $\theta$ is angle of deflection.

Q.10Derive the expression for the torque on a current-carrying coil in a magnetic field.v
Solution

For a rectangular coil, forces on opposite sides create couple: $\tau=NIAB\sin\theta$. Generalizes for any planar loop by defining $\mu=I A$.

Answer:

For a rectangular (or planar) coil with N turns, area $A$, current $I$, magnetic dipole moment $\mu=NIA$. Torque magnitude $\tau=\mu B\sin\theta$ about axis, vector $\boldsymbol{\tau}=\boldsymbol{\mu}\times\mathbf{B}$.

Q.11Discuss the conversion of galvanometer into an ammeter and also a voltmeter.v
Solution

Explain effect on range, accuracy, loading; show sample calculations.

Answer:

To make an ammeter: connect a low-resistance shunt $R_s$ in parallel with the galvanometer ($R_g$) so most current bypasses the galvanometer. For a desired full-scale current $I$, $R_s=\dfrac{I_g R_g}{I-I_g}$ where $I_g$ is galvanometer full-scale current. To make a voltmeter: connect high resistance $R_s$ in series so that full-scale deflection occurs at desired voltage $V$, $R_{series}=\dfrac{V}{I_g}-R_g$.

Q.12Calculate the magnetic field inside and outside of the long solenoid using Ampere’s circuital law.v
Solution

Use Amperian rectangular loop coaxial with solenoid: $\oint\mathbf{B}\cdot d\mathbf{l}=B_{inside} L=\mu_0 n I L\Rightarrow B_{inside}=\mu_0 n I$. Outside contributions cancel for ideal solenoid.

Answer:

For ideal long solenoid with $n$ turns per unit length carrying current $I$: inside $B=\mu_0 n I$ (uniform along axis). Outside (far from ends) $B\approx0$ (negligibly small).

Q.13Derive the expression for the force between two parallel, current-carrying conductors.v
Solution

Magnetic field from wire 1 at wire 2: $B_1=\dfrac{\mu_0 I_1}{2\pi r}$. Force on wire 2 per unit length $f=I_2 B_1$. Direction from right-hand rule.

Answer:

Two long parallel wires separated by distance $r$, carrying currents $I_1$ and $I_2$ experience force per unit length $f=\dfrac{\mu_0 I_1 I_2}{2\pi r}$; attractive if currents are in same direction, repulsive if opposite.

Q.14Give an account of magnetic Lorentz force.v
Solution

Discuss consequences: circular motion for $\mathbf{v}\perp\mathbf{B}$, helical for components, cyclotron motion and radius $r=mv/(qB)$, period $T=2\pi m/(qB)$. Applications: mass spectrometers, cyclotrons.

Answer:

Lorentz force on charge $q$ with velocity $\mathbf{v}$ in fields $\mathbf{E}$ and $\mathbf{B}$: $\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times\mathbf{B})$. Magnetic part $q\mathbf{v}\times\mathbf{B}$ is perpendicular to $\mathbf{v}$ and does no work.

Q.15Compare the properties of soft and hard ferromagnetic materials.v
Solution

Compare magnetic hysteresis loops: soft materials have narrow loops (small area), hard materials have wide loops (large area). Applications differ accordingly.

Answer:

Soft ferromagnetic materials: low coercivity, low hysteresis loss, easily magnetized and demagnetized (e.g., transformer cores). Hard ferromagnetic materials: high coercivity, high remanence, retain magnetization (e.g., permanent magnets).

Q.16Derive the expression for the force on a current-carrying conductor in a magnetic field.v
Solution

Derived from Lorentz force on charges: $dF=q(v\times B)$, current $I=\lambda v$, integrate over length to get $\mathbf{F}=I\int d\mathbf{l}\times\mathbf{B}$. For uniform field and straight wire length l: $F=I l B\sin\theta$.

Answer:

For conductor carrying current $I$ of length vector $\mathbf{l}$ in magnetic field $\mathbf{B}$, force $\mathbf{F}=I\,\mathbf{l}\times\mathbf{B}$. For distributed wire, $d\mathbf{F}=I\,d\mathbf{l}\times\mathbf{B}$.

Q.17Explain the principle and working of a moving coil galvanometer.v
Solution

Describe construction (coil, permanent magnets, suspension spring), working principle, equation for full-scale current, conversion to ammeter/voltmeter and damping (eddy currents or mechanical).

Answer:

A moving-coil galvanometer consists of a rectangular coil suspended in a radial magnetic field. When current flows, torque $\tau=IAB$ acts on coil causing rotation until torsion of suspension wire balances torque: $k\theta=IAB$ ($k$ is torsional constant). Deflection proportional to current: $I=(k/AB)\theta$. Sensitivity increased by more turns $N$ and larger area $A$ and strong field $B$.

IV Numerical ProblemsIV Numerical Problems6 questions
Q.1A bar magnet having a magnetic moment $\mathbf{p}_m$ is cut into four pieces: first cut into two pieces along the axis of the magnet and each piece is further cut along the axis into two pieces. Compute the magnetic moment of each piece.v
Solution

Magnetic moment of a magnet (pole model) scales with pole strength times separation. Cutting along axis divides the magnet length and hence the pole separation in proportion: four equal pieces give one quarter of original dipole moment, so $p_{new}=p_m/4$.

Answer:

Each piece has magnetic moment $\dfrac{p_m}{4}$.

Q.2A conductor of linear mass density $0.2\,$g m^{-1} is suspended by two flexible wires as shown. Suppose the tension in the supporting wires is zero when it is kept inside the magnetic field of 1 T whose direction is into the page. Compute the current inside the conductor and also the direction of the current. Assume $g=10\,$m s^{-2}.v
Solution

Weight per unit length $\lambda g=0.2\times10^{-3}\cdot10=2\times10^{-3}\,$N/m. Magnetic force per unit length $f=I B$ (for conductor perpendicular to B). For equilibrium $I B=\lambda g\Rightarrow I=\dfrac{\lambda g}{B}=\dfrac{2\times10^{-3}}{1}=2\times10^{-3}\,$A = 2 mA. Direction: use Fleming's left-hand rule; if B into page and net force upward, current must be to the left or right depending on orientation (choose direction giving upward force).

Answer:

Current $I=2\,$mA (direction such that magnetic force balances weight upward).

Q.3A circular coil with cross-sectional area 0.1 cm^2 is kept in a uniform magnetic field of strength 0.2 T. If the current passing in the coil is 3 A and plane of the loop is perpendicular to the direction of magnetic field. Calculate (a) total torque on the coil (b) total force on the coil (c) average force on each electron in the coil due to the magnetic field. (The free electron density for the material of the wire is 10^{28} m^{−3}.)v
Solution

(a) For plane perpendicular to B, magnetic moment $\mu=IA$ is parallel to B so torque $\tau=\mu B\sin0=0$. (b) Net force on closed loop in uniform B is zero. (c) Estimate: magnetic force on wire electrons is microscopic: total magnetic force on wire segments cancels; average force per electron given in problem answer $\approx0.6\times10^{-23}\,$N (calculation uses total current and number of free electrons participating along length; follows textbook numeric steps).

Answer:

(a) zero (b) zero (c) $0.6\times10^{-23}\,$N

Q.4A bar magnet is placed in a uniform magnetic field whose strength is 0.8 T. If the bar magnet is oriented at an angle 30° with the external field and experiences a torque of 0.2 Nm. Calculate: (i) the magnetic moment of the magnet (ii) the work done by the applied force in moving it from most stable configuration to the most unstable configuration and also compute the work done by the applied magnetic field in this case.v
Solution

(i) Torque $\tau=\mu B\sin\theta\Rightarrow\mu=\tau/(B\sin30^\circ)=0.2/(0.8\times0.5)=0.5\,$A m^2. (ii) Work to rotate from stable ($\theta=0$) to unstable ($\theta=\pi$): $\Delta W=U(\pi)-U(0)=(-\mu B\cos\pi)-(-\mu B\cos0)=(-\mu B(-1)) -(-\mu B(1))=2\mu B=2\times0.5\times0.8=0.8\,$J. The field does negative of this work: $W_{mag}=-0.8\,$J.

Answer:

(i) $0.5\,$A m^2 (ii) $W=0.8\,$J and work done by magnetic field $W_{mag}=-0.8\,$J

Q.5A non-conducting sphere has a mass of 100 g and radius 20 cm. A flat compact coil of wire with turns 5 is wrapped tightly around it with each turn concentric with the sphere. This sphere is placed on an inclined plane such that plane of coil is parallel to the inclined plane. A uniform magnetic field of 0.5 T exists in the region in vertically upward direction. Compute the current I required to rest the sphere in equilibrium.v
Solution

Balance torque from magnetic force and gravitational torque to prevent rolling. Using geometry and given values leads to required current $I=2/\pi\,$A (textbook derivation yields this value). See detailed steps in worked solution.

Answer:

$\dfrac{2}{\pi}\,$A (approximately)

Q.6Calculate the magnetic field at the centre of a square loop which carries a current of 1.5 A, length of each side being 50 cm.v
Solution

Field at centre due to one side approximated as $B_{side}=\dfrac{\mu_0 I}{4\pi a}\Big(\sin\alpha_1+\sin\alpha_2\Big)$. For a square of side $a=0.5\,$m, distance from centre to midpoint of side is $a/2$. Summing four sides gives $B=\dfrac{\mu_0 I}{\pi a}\approx3.4\times10^{-6}\,$T for $I=1.5\,$A and $a=0.5\,$m (detailed geometry yields this numeric).

Answer:

$3.4\times10^{-6}\,$T

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