Q.A toroid of $n$ turns, mean radius $R$ and cross-sectional radius $a$ carries current $I$. It is placed on a horizontal table taken as $x$-$y$ plane. Its magnetic moment $m$

Showing the 12 most recent of 16 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.On both sides of a magnetic needle, two short magnets A and B are placed on the same horizontal line which is perpendicular to the magnetic meridian. The south poles of $A$ and $B$ are facing each other, which are 10 cm and 20 cm respectively from the magnetic needle. If the needle remains undeflected, the ratio of the magnetic moment of $A$ to that $B$ is: (A) $2:1$ (B) $8:1$ (C) $1: 8$ (D) $1: 2$

   ›Reveal solutionSolution

   The needle stays undeflected when the net magnetic field from the two magnets at the needle’s location is zero. Using the axial field formula for a short magnet, the ratio of magnetic moments is found to be $1:8$, so the correct option is (C).

   Concept & Intuition

   A magnetic needle aligns with the local magnetic field. Here, two short magnets are placed on the same line perpendicular to the magnetic meridian, with their south poles facing each other. The needle is between them. For it to remain undeflected, the magnetic fields produced by A and B at the needle’s position must cancel exactly. Since both magnets are “short,” we can treat them as magnetic dipoles and use the axial field formula $B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3}$ for a point on the axis of a dipole. The distances from the needle are given, so we set the field magnitudes equal and solve for the ratio of magnetic moments.

   Step-by-step reasoning

   1. Identify the field direction

      Both magnets have their south poles facing the needle. On the axis of a short magnet, the field points away from the north pole and toward the south pole. Since the south poles are closer to the needle, the field from each magnet at the needle points toward that magnet. Thus, the fields from A and B are in opposite directions (one left, one right), so they can cancel.

   2. Write the axial field formula

      For a short magnet of magnetic moment $M$, the magnetic field at a distance $r$ along its axis (in vacuum) is

   $B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3}.$

   This holds when $r$ is much larger than the magnet’s length — given as “short magnets,” this is valid.

   3. Set up the cancellation condition Let $M_A$ and $M_B$ be the magnetic moments. Distances from the needle: $r_A = 10\ \text{cm}$, $r_B = 20\ \text{cm}$. For no deflection,

   $B_A = B_B \quad \Rightarrow \quad \frac{2M_A}{r_A^3} = \frac{2M_B}{r_B^3}.$

   The factor $\frac{\mu_0}{4\pi}$ cancels.

   4. Solve for the ratio Cancel the factor 2:

   $\frac{M_A}{r_A^3} = \frac{M_B}{r_B^3} \quad \Rightarrow \quad \frac{M_A}{M_B} = \frac{r_A^3}{r_B^3}.$

   Substitute $r_A = 10$, $r_B = 20$:

   $\frac{M_A}{M_B} = \frac{10^3}{20^3} = \frac{1000}{8000} = \frac{1}{8}.$

   So the ratio $M_A : M_B = 1 : 8$.

   Tip

   Notice that the ratio depends on the cube of the distances — a small change in distance greatly affects the field. Here, doubling the distance reduces the field by a factor of 8, so the farther magnet must have 8 times the moment to balance.

   Watch out

   A common mistake is to use the inverse-square law (like for electric charges) instead of the inverse-cube law for dipoles. Always check: a magnetic dipole’s field falls off as $1/r^3$, not $1/r^2$.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

2. COMEDK 2026Set 2026-M1 markMCQ
   Q.A bar magnet of length 12 cm is placed such that its north pole points towards the geographic north. Two neutral points which are separated by 16 cm are obtained on the equatorial axis of the bar magnet. What is the pole strength of the bar magnet if the horizontal component of the earth's field is $1.25 \times 10^{-5} \mathrm{~T}$ ? (A) 3.024 Am (B) 1.042 Am (C) 10.24 Am (D) 2.032 Am

   ›Reveal solutionSolution

   With the north pole pointing geographic north, the neutral points lie on the equatorial line, where the magnet's field cancels the earth's horizontal component $B_H$. With half-length $l=6$ cm and $r=8$ cm, equating $B_{\text{eq}}=B_H$ gives pole strength $m\approx 1.042\ \text{A m}$ — option (B).

   Concept. North pole toward geographic north $\Rightarrow$ neutral points are broadside-on (equatorial). There the magnet's equatorial field is antiparallel to $B_H$ and equal in magnitude.

   Step 1 — Geometry. Length $=12$ cm $\Rightarrow l=6\ \text{cm}=0.06\ \text{m}$. The two neutral points are $16$ cm apart and symmetric about the centre, so each is at

   $r=\tfrac{16}{2}=8\ \text{cm}=0.08\ \text{m}.$

   Step 2 — Equatorial field of the magnet.

   $B_{\text{eq}}=\frac{\mu_0}{4\pi}\cdot\frac{M}{(r^2+l^2)^{3/2}},\qquad M=m\cdot(2l)=m\times 0.12.$

   Step 3 — Neutral-point condition $B_{\text{eq}}=B_H$.

   $r^2+l^2=(0.08)^2+(0.06)^2=0.0064+0.0036=0.01\ \text{m}^2,$

   $(r^2+l^2)^{3/2}=(0.01)^{3/2}=0.001\ \text{m}^3.$

   $10^{-7}\cdot\frac{0.12\,m}{0.001}=1.25\times10^{-5}.$

   Step 4 — Solve.

   $10^{-7}\times 120\,m=1.25\times10^{-5}\ \Rightarrow\ 1.2\times10^{-5}\,m=1.25\times10^{-5},$

   $m=\frac{1.25}{1.2}\approx 1.042\ \text{A m}.$

   ✓Final answer

   Pole strength $m\approx 1.042\ \text{A m}$ — option (B).

   ANSWER: B

3. KCET 2025Set D-41 markMCQ
   Q.Two thin long parallel wires separated by a distance $r$ from each other in vacuum carry a current of $I$ ampere in opposite directions. Then, they will (A) Attract each other with a force per unit length of $\dfrac{\mu_0 I^2}{2\pi r}$ (B) Repel each other with a force per unit length of $\dfrac{\mu_0 I^2}{2\pi r}$ (C) Repel each other with a force per unit length of $\dfrac{\mu_0 I^2}{2\pi r^2}$ (D) Attract each other with a force per unit length of $\dfrac{\mu_0 I^2}{2\pi r^2}$

   ›Reveal solutionSolution

   Antiparallel currents repel; the force per unit length comes from $B = \mu_0 I/2\pi r$ combined with $F = BIL$, giving $F/L = \mu_0 I^2 / 2\pi r$.

   Step 1 — The field of the first wire.

   A long straight wire carrying current $I_1$ produces, at perpendicular distance $r$, a magnetic field of magnitude

   $B_1 = \frac{\mu_0 I_1}{2\pi r}$

   circling the wire (right-hand rule). At the position of the second wire, this field is perpendicular to that wire.

   Step 2 — The force on the second wire.

   A length $L$ of the second wire, carrying $I_2$ perpendicular to $\vec{B}_1$, feels

   $F = B_1 I_2 L \sin 90^\circ = B_1 I_2 L$

   Substituting $B_1$:

   $F = \frac{\mu_0 I_1 I_2 L}{2\pi r} \quad\Longrightarrow\quad \boxed{\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}}$

   With both currents equal to $I$ as stated:

   $\frac{F}{L} = \frac{\mu_0 I^2}{2\pi r}$

   Step 3 — The direction: attract or repel?

   Apply the right-hand rule twice.

   Take wire 1 carrying current up the page and wire 2, a distance $r$ to its right, carrying current down the page (opposite directions).

   • Wire 1's field at wire 2's location points into the page (curl the right hand around wire 1).
   • The force on wire 2 is $\vec{F} = I_2 \vec{L} \times \vec{B}_1$, with $\vec{L}$ pointing down and $\vec{B}_1$ into the page. Then $(\text{down}) \times (\text{into page})$ points to the right — i.e. away from wire 1.

   By Newton's third law wire 1 is pushed equally to the left, away from wire 2. So the wires repel.

   The general rule, worth memorising:

   $\textbf{Currents in the SAME direction} \Rightarrow \textbf{ATTRACT}$

   $\textbf{Currents in OPPOSITE directions} \Rightarrow \textbf{REPEL}$

   Since the question states the currents are opposite, the wires repel. This eliminates (A) and (D).

   Step 4 — Choose between (B) and (C): the power of $r$.

   Both remaining options say "repel", so only the formula distinguishes them. From Step 2 the dependence is

   $\frac{F}{L} \propto \frac{1}{r}$

   — a single power of $r$, because it inherits the $1/r$ falloff of a long straight wire's magnetic field. Option (C)'s $1/r^2$ would be dimensionally wrong for a force per unit length here and confuses this with an inverse-square law. So (B) is correct.

   Aside: this very formula, with $I = 1\ \text{A}$ and $r = 1\ \text{m}$ giving $F/L = 2\times10^{-7}\ \text{N m}^{-1}$, was historically the definition of the ampere.

   ✓Final answer

   The correct option is (B) — Repel each other with a force per unit length of $\dfrac{\mu_0 I^2}{2\pi r}$.

   ANSWER: B

4. KCET 2025Set D-41 markMCQ
   Q.Identify the correct statement (A) A current carrying conductor produces an electric field around it. (B) A straight current carrying conductor has circular magnetic field lines around it. (C) The direction of magnetic field due to a current element is given by Flemings Left Hand Rule (D) The magnetic field inside a solenoid is non-uniform

   ›Reveal solutionSolution

   Test each statement against the Biot–Savart law and the right-hand rule; only the "circular field lines around a straight wire" statement survives.

   Step 1 — Option (B): the correct statement.

   For a long straight wire, the Biot–Savart law integrates to

   $B = \frac{\mu_0 I}{2\pi r}$

   This depends only on $r$, the perpendicular distance from the wire. So every point at the same distance $r$ has the same field magnitude — the lines of constant $B$ are circles centred on the wire.

   The direction follows from the right-hand thumb rule: point the right thumb along the current, and the curled fingers give the sense of $\vec{B}$. That curl is exactly a circle, in a plane perpendicular to the wire. The field lines are therefore closed concentric circles — consistent with $\nabla\cdot\vec{B} = 0$ (magnetic field lines never begin or end, since magnetic monopoles do not exist). ✓ True.

   Step 2 — Option (A): "produces an electric field around it." ✗

   A steady current-carrying conductor is electrically neutral — the drifting electrons' charge is exactly balanced by the fixed positive lattice ions. With zero net charge density, there is no external electrostatic field. What it does produce is a magnetic field. (There is an electric field inside the wire, driving the current, but that is not "around it".)

   Step 3 — Option (C): "direction of magnetic field due to a current element is given by Fleming's Left Hand Rule." ✗

   This confuses two different rules:

   • Fleming's Left Hand Rule gives the direction of the force on a current-carrying conductor placed in an external magnetic field ($\vec{F} = I\vec{L}\times\vec{B}$). It is a rule about force, not about field.
   • The direction of the field produced by a current element comes from the Biot–Savart law,

   $d\vec{B} = \frac{\mu_0}{4\pi}\frac{I\, d\vec{l} \times \hat{r}}{r^2}$

   whose cross product is evaluated with the right-hand rule.

   So the statement names the wrong rule for the wrong quantity, with the wrong hand.

   Step 4 — Option (D): "the magnetic field inside a solenoid is non-uniform." ✗

   Ampère's law applied to a long solenoid gives

   $B = \mu_0 n I$

   which contains no positional variable — the field is the same everywhere well inside the solenoid, both along the axis and off it. Uniformity is precisely why a solenoid is the standard laboratory source of a uniform magnetic field. (Only near the open ends does the field weaken and fringe — falling to about $\tfrac{1}{2}\mu_0 n I$ right at the mouth — but the statement, made without that qualification, is false.)

   Step 5 — Conclusion.

   Three statements fail on definite physical grounds; only (B) is correct.

   ✓Final answer

   The correct option is (B) — A straight current carrying conductor has circular magnetic field lines around it.

   ANSWER: B

5. KCET 2025Set D-41 markMCQ
   Q.When a bar magnet is pushed towards the coil, along its axis, as shown in the figure, the galvanometer pointer deflects towards X. When this magnet is pulled away from the coil, the galvanometer pointer (A) Deflects towards X' (B) Does not deflect (C) Oscillates (D) Deflects towards X

   ›Reveal solutionSolution

   Reversing the direction of the flux change reverses the sign of the induced emf (Lenz's law), so the pointer swings the other way — towards X'.

   Step 1 — The governing law.

   Faraday's law of electromagnetic induction with Lenz's sign convention:

   $\varepsilon = -\frac{d\Phi_B}{dt}$

   The minus sign is Lenz's law: the induced current flows in whatever sense opposes the change in flux that caused it. It is a direct statement of conservation of energy — if the current helped the change, the magnet would accelerate on its own and give energy for free.

   Step 2 — What happens when the magnet is pushed towards the coil.

   As the magnet approaches, the flux linked with the coil increases. The induced current therefore circulates so as to oppose the approach — the face of the coil towards the magnet acquires the same polarity as the approaching pole and repels it. Some definite current direction results, and we are told the galvanometer reads this as a deflection towards X.

   Step 3 — What happens when the magnet is pulled away.

   Now the flux linked with the coil decreases. Lenz's law again demands opposition, but the change to be opposed is the opposite one: the coil now tries to hold the magnet back, so the face towards the magnet takes the opposite polarity and attracts it.

   Since the sign of $d\Phi_B/dt$ has flipped from positive to negative, the sign of $\varepsilon$ flips too, and the induced current reverses direction in the coil and hence through the galvanometer.

   Step 4 — Read the galvanometer.

   A galvanometer is a centre-zero, direction-sensitive instrument: reversing the current through it reverses the side to which the pointer swings. Pushing in $\to$ X; therefore pulling out $\to$ the other side, X'.

   Step 5 — Reject the rest.

   • (B) "Does not deflect" — wrong: the flux is still changing, so an emf and a current definitely exist.
   • (C) "Oscillates" — wrong: the flux changes steadily in one sense during withdrawal, giving a steady one-way current, not an alternating one.
   • (D) "Deflects towards X" — wrong: that would mean the current direction is unchanged, which contradicts Lenz's law.

   ✓Final answer

   The correct option is (A) — Deflects towards X' (the current reverses, because the flux is now decreasing instead of increasing).

   ANSWER: A

6. COMEDK 2025Set 2025-A1 markMCQ
   Q.For a short magnet, the magnetic field on the axial line at a distance 10 cm from its centre is $1.6 \mathrm{~m} \times 10^{-7} \mathrm{~T}$. What is the magnetic field on its equatorial line at the same distance 10 cm from its centre? (A) $4.8 \times 10^{-7} T$ (B) $3.2 \times 10^{-8} T$ (C) $1.6 \times 10^{-7} T$ (D) $0.8 \times 10^{-7} T$

   ›Reveal solutionSolution

   For a short bar magnet, the axial field is twice the equatorial field at the same distance. Given axial field = $1.6 \times 10^{-7} \, \mathrm{T}$, the equatorial field is $0.8 \times 10^{-7} \, \mathrm{T}$, so the correct option is (D).

   Concept & Intuition

   A short bar magnet produces a magnetic field that is not isotropic — it is stronger along the axis (the line through the north and south poles) than along the equatorial line (the perpendicular bisector). For a point at the same distance from the centre, the axial field is exactly twice the equatorial field. This comes from the geometry of the dipole field: on the axis, the contributions from both poles add constructively; on the equator, they partially cancel. So if you know one, you immediately know the other — no need to re-derive from scratch.

   Step-by-step reasoning

   1. Recall the standard formulas for a short bar magnet For a magnetic dipole of moment $M$, at a distance $r$ from the centre (where $r$ is much larger than the magnet’s length), the field magnitudes are:
      • On the axial line:

   $B_{\text{axial}} = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3}$

   • On the equatorial line:

   $B_{\text{equatorial}} = \frac{\mu_0}{4\pi} \cdot \frac{M}{r^3}$

   2. Compare the two expressions Dividing the axial by the equatorial:

   $\frac{B_{\text{axial}}}{B_{\text{equatorial}}} = \frac{2M/r^3}{M/r^3} = 2$

   Hence,

   $B_{\text{axial}} = 2 \cdot B_{\text{equatorial}}$

   or equivalently,

   $B_{\text{equatorial}} = \frac{1}{2} B_{\text{axial}}$

   3. Apply the given data The axial field at 10 cm is $1.6 \times 10^{-7} \, \mathrm{T}$. Therefore:

   $B_{\text{equatorial}} = \frac{1}{2} \times 1.6 \times 10^{-7} = 0.8 \times 10^{-7} \, \mathrm{T}$

   4. Match with the options The value $0.8 \times 10^{-7} \, \mathrm{T}$ corresponds to option (D).

   Watch out

   A common mistake is to think the equatorial field is zero or to confuse the factor — some students recall the electric dipole analogy incorrectly. For a magnetic dipole, the axial field is double the equatorial field, not equal or half.

   Tip

   You don’t need to compute $\mu_0$ or $M$ at all. The ratio is fixed for a short dipole, so the problem reduces to a simple halving.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

7. KCET 2023Set A-31 markMCQ
   Q.The torque acting on a magnetic dipole placed in uniform magnetic field is zero, when the angle between the dipole axis and the magnetic field is (A) $45^\circ$ (B) $60^\circ$ (C) $90^\circ$ (D) zero

   ›Reveal solutionSolution

   $\vec\tau = \vec m \times \vec B$, so $\tau = mB\sin\theta = 0$ when $\theta = 0$.

   Step 1 — The torque law.

   A magnetic dipole of moment $\vec m$ in a uniform field $\vec B$ feels no net force, but it does feel a couple:

   $\vec\tau = \vec m \times \vec B, \qquad |\vec\tau| = mB\sin\theta,$

   where $\theta$ is the angle between the dipole axis $\vec m$ and $\vec B$. The cross product is what makes the torque vanish when the two vectors are parallel.

   Step 2 — Set the torque to zero.

   $mB\sin\theta = 0 \;\Longrightarrow\; \sin\theta = 0 \;\Longrightarrow\; \theta = 0^\circ \text{ (or } 180^\circ).$

   Step 3 — Check the given options.

   $\theta = 45^\circ:\ \tau = \tfrac{1}{\sqrt2}mB \ne 0; \qquad \theta = 60^\circ:\ \tau = \tfrac{\sqrt3}{2}mB \ne 0;$

   $\theta = 90^\circ:\ \tau = mB \;(\textbf{maximum, not zero}); \qquad \theta = 0^\circ:\ \tau = 0. \;\checkmark$

   Physically: when the dipole is already aligned with the field it is in its stable equilibrium position — there is nothing left to rotate it. The $90^\circ$ option is the classic trap, because that is where the torque is largest.

   ✓Final answer

   The correct option is (D) — zero.

   ANSWER: D

8. COMEDK 2023Set 2023-E1 markMCQ
   Q.Find the pole strength of a magnet of length $2 \mathrm{~cm}$, if the magnetic field strength $B$ at distance $10 \mathrm{~cm}$ from the centre of a magnet on the axial line of the magnet is $10^{-4} \mathrm{~T}$. (A) $25 \mathrm{~Am}$ (B) $100 \mathrm{~Am}$ (C) $5 \times 10^{-2} \mathrm{~Am}$ (D) $1 \times 10^{-4} \mathrm{~Am}$

   ›Reveal solutionSolution

   Using the axial field of a bar magnet, $B=\dfrac{\mu_0}{4\pi}\dfrac{2M}{d^3}$ with $M=m\cdot2l$, the pole strength comes out to $m\approx25$ Am.

   Magnetic moment $M=m\times(2l)$, where $2l=2$ cm $=0.02$ m is the magnet length and $m$ the pole strength. Distance $d=10$ cm $=0.1$ m.

   For a short magnet, axial field:

   $B=\frac{\mu_0}{4\pi}\frac{2M}{d^3}=10^{-7}\cdot\frac{2\,m(0.02)}{(0.1)^3}.$

   Set $B=10^{-4}$ T:

   $10^{-4}=10^{-7}\cdot\frac{0.04\,m}{10^{-3}}=10^{-7}\cdot40\,m=4\times10^{-6}\,m.$

   $m=\frac{10^{-4}}{4\times10^{-6}}=25\ \text{Am}.$

   (Using the exact axial formula $B=\dfrac{\mu_0}{4\pi}\dfrac{2Md}{(d^2-l^2)^2}$ gives essentially the same $m\approx25$ Am.)

   ✓Final answer

   The correct option is (A) — $25\,\text{Am}$

9. COMEDK 2023Set 2023-M1 markMCQ
   Q.If $\theta_1$ and $\theta_2$ be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of $\operatorname{dip} \theta$ is given by (A) $\cot ^2 \theta=\cot ^2 \theta_1+\cot ^2 \theta_2$ (B) $\tan ^2 \theta=\tan ^2 \theta_1+\tan ^2 \theta_2$ (C) $\cot ^2 \theta=\cot ^2 \theta_1-\cot ^2 \theta_2$ (D) $\tan ^2 \theta=\tan ^2 \theta_1-\tan ^2 \theta_2$

   ›Reveal solutionSolution

   The true dip relates to the two perpendicular apparent dips by $\cot^2\theta=\cot^2\theta_1+\cot^2\theta_2$.

   Let the total field be $B$, its horizontal component $B_H$ and vertical component $B_V$, so $\tan\theta=B_V/B_H$. In a vertical plane making angle $\alpha$ with the magnetic meridian, the effective horizontal component is $B_H\cos\alpha$, while the vertical component is unchanged, giving apparent dip

   $\tan\theta_1=\frac{B_V}{B_H\cos\alpha},\qquad \tan\theta_2=\frac{B_V}{B_H\cos(90^\circ-\alpha)}=\frac{B_V}{B_H\sin\alpha}.$

   Hence

   $\cot^2\theta_1+\cot^2\theta_2=\frac{B_H^2\cos^2\alpha}{B_V^2}+\frac{B_H^2\sin^2\alpha}{B_V^2}=\frac{B_H^2}{B_V^2}=\cot^2\theta.$

   ✓Final answer

   The correct option is (A) — $\cot^2\theta=\cot^2\theta_1+\cot^2\theta_2$

10. COMEDK 2022Set 20221 markMCQ
    Q.A bar magnet of length $6 \mathrm{~cm}$, is placed in the magnetic meridian with $N$ pole, pointing towards the geographical north. Two neutral points, separated by a distance of $8 \mathrm{~cm}$ are obtained on the equitorial axis of the magnet. If $B_H=1.2 \times 10^{-5} \mathrm{~T}$. Then the pole strength of the magnet is (A) 0.75 A-m$^{2}$ (B) 0.25 A-m$^{2}$ (C) 0.50 A-m$^{2}$ (D) 1.50 A-m$^{2}$

    ›Reveal solutionSolution

    (The option's unit A-m^2 is a misprint; the numerical value is 0.25.)

    Concept: neutral points occur where the magnet's field exactly cancels the horizontal component of the Earth's field. With the N pole pointing geographic north, the neutral points lie on the EQUATORIAL line of the magnet.

    Geometry:

    magnet length 2l = 6 cm -> l = 3 cm = 0.03 m

    the two neutral points are 8 cm apart, symmetric about the centre -> d = 4 cm = 0.04 m

    Equatorial field of a bar magnet:

    B = (mu0/4pi) M / (d^2 + l^2)^(3/2) = B_H

    d^2 + l^2 = (0.04)^2 + (0.03)^2 = 0.0016 + 0.0009 = 0.0025 m^2

    (d^2 + l^2)^(3/2) = (0.05)^3 = 1.25 x 10^-4 m^3

    10^-7 x M / (1.25 x 10^-4) = 1.2 x 10^-5

    M = 1.2 x 10^-5 x 1.25 x 10^-4 / 10^-7 = 1.5 x 10^-2 A m^2

    Pole strength:

    m = M / (2l) = 1.5 x 10^-2 / 0.06 = 0.25 A m

    (The option's unit A-m^2 is a misprint; the numerical value is 0.25.)

    ✓Final answer

    The correct option is (B) — 0.25 A-m$^{2}$

    ANSWER: B

11. KCET 2021Set B-21 markMCQ
    Q.A toroid with thick windings of $N$ turns has inner and outer radii $R_1$ and $R_2$ respectively. If it carries certain steady current $I$, the variation of the magnetic field due to the toroid with radial distance is correctly graphed in (A) (B) (C) (D)

    ›Reveal solutionSolution

    Apply Ampère's law to the toroid: $B$ is non-zero only inside the core, where $B=\mu_0 NI/2\pi r$; at the mean radius this is the constant $\mu_0 NI/\pi(R_1+R_2)$ — the flat plateau of graph (D).

    Step 1 — Where the field exists. Choose a circular Amperian loop of radius $r$ concentric with the toroid. For $r<R_1$ (the central hole) the loop encloses no current, so $B=0$. For $r>R_2$ (outside) each winding's current is enclosed once going in and once coming out, so the net enclosed current is again zero and $B=0$. Only for $R_1<r<R_2$ (inside the core) does the loop enclose the full $NI$.

    Step 2 — The field inside the core. Ampère's law on that loop gives $B\,(2\pi r)=\mu_0 NI$, so $B=\dfrac{\mu_0 NI}{2\pi r}$.

    Step 3 — The mean-radius value. For a toroid the field is customarily quoted at the mean radius $\bar r=\dfrac{R_1+R_2}{2}$:

    $B=\frac{\mu_0 NI}{2\pi\bar r}=\frac{\mu_0 NI}{\pi(R_1+R_2)}$

    Graph (D) plots exactly this: $B$ constant at $\dfrac{\mu_0 NI}{\pi(R_1+R_2)}$ across the core band and zero on either side.

    Why the other graphs are wrong. (A) and (B) show $B$ rising from $r=0$ — but the field in the central hole is exactly zero (no enclosed current), and (A)'s printed levels $\mu_0 NI/2R$ even omit the $\pi$. (C) shows $B$ increasing from $R_1$ to $R_2$ — but $B\propto 1/r$ can only decrease with $r$.

    (Strictly, $B$ falls as $1/r$ from $\mu_0 NI/2\pi R_1$ to $\mu_0 NI/2\pi R_2$ across the core; the official answer uses the standard mean-radius idealisation, which treats $B$ as uniform at its mean value — the only option drawn to that model, at the exact printed value.)

    ✓Final answer

    (D)

12. KCET 2021Set B-21 markMCQ
    Q.Earth’s magnetic field always has a horizontal component except at (A) equator (B) magnetic poles (C) a latitude of $60^\circ$ (D) an altitude of $60^\circ$

    ›Reveal solutionSolution

    $H = B\cos\delta$ is zero only when the dip $\delta = 90^\circ$, which occurs at the magnetic poles where the Earth's field is purely vertical.

    1. Resolve the Earth's field into components

    At any place on the Earth the total field $\vec B$ makes an angle $\delta$ (the angle of dip or inclination) with the horizontal. Resolving:

    $H = B\cos\delta \qquad\text{(horizontal component)}$

    $V = B\sin\delta \qquad\text{(vertical component)}$

    with

    $\tan\delta = \frac{V}{H}, \qquad B=\sqrt{H^2+V^2}$

    2. When can $H$ be zero?

    Since $B \neq 0$ anywhere on Earth,

    $H = 0 \iff \cos\delta = 0 \iff \delta = 90^\circ$

    A dip of $90^\circ$ means the field is entirely vertical — the needle of a dip circle stands straight up. This is precisely the definition of the magnetic poles of the Earth. (There, a freely suspended compass needle has no horizontal direction to point in, which is why an ordinary compass is useless at the magnetic poles.)

    3. Check the other options

    1. (A) Equator — at the magnetic equator $\delta = 0^\circ$, so $H = B\cos 0^\circ = B$ and $V = 0$. The field is entirely horizontal: this is the place where $H$ is maximum, the exact opposite of what is asked. ✗
    2. (C) A latitude of $60^\circ$ — using the dipole relation $\tan\delta = 2\tan\lambda$, a magnetic latitude $\lambda = 60^\circ$ gives $\tan\delta = 2\tan 60^\circ = 3.46$, so $\delta \approx 74^\circ$ — large, but $\cos 74^\circ \neq 0$, so $H$ is small yet non-zero. ✗
    3. (D) An altitude of $60^\circ$ — "altitude" is not even a meaningful parameter for the dip; changing height above the ground does not tilt the field to vertical. ✗

    ✓Final answer

    The correct option is (B) — magnetic poles.

    ANSWER: B