Q.Two identical current carrying coaxial loops, carry current $I$ in an opposite sense. A simple amperian loop passes through both of them once. Calling the loop as $C$,

Showing the 12 most recent of 17 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Two long straight parallel wires P and Q carrying currents of $10\,\mathrm{A}$ and $20\,\mathrm{A}$ respectively in the same direction are placed in air with a separation of $10\,\mathrm{cm}$ between them. Another long straight wire R carrying a current of $5\,\mathrm{A}$ in the opposite direction is placed between the wires P and Q and parallel to them at a distance of $5\,\mathrm{cm}$ from wire Q. Then the magnitude of the net force acting on wire R per unit length is (A) $6 \times 10^{-4}\ \mathrm{Nm}^{-1}$ (B) $2 \times 10^{-4}\ \mathrm{Nm}^{-1}$ (C) $4 \times 10^{-4}\ \mathrm{Nm}^{-1}$ (D) $8 \times 10^{-4}\ \mathrm{Nm}^{-1}$

   ›Reveal solutionSolution

   R sits midway between P and Q. Both P and Q repel R (opposite currents), but from opposite sides, so the forces subtract: net $=4\times10^{-4}-2\times10^{-4}=2\times10^{-4}~\text{N m}^{-1}$.

   Setup

   • Wire P: $I_P = 10~\text{A}$; Wire Q: $I_Q = 20~\text{A}$ (same direction), separation $10~\text{cm}$.
   • Wire R: $I_R = 5~\text{A}$ in the opposite direction, placed between them at $5~\text{cm}$ from Q — hence also $5~\text{cm}$ from P (exactly midway).

   Force per unit length between two parallel wires:

   $\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}, \qquad \frac{\mu_0}{2\pi} = 2\times10^{-7}$

   Force on R due to P ($d = 0.05~\text{m}$):

   $\frac{F_P}{L} = \frac{(2\times10^{-7})(10)(5)}{0.05} = 2\times10^{-4}~\text{N m}^{-1}$

   Opposite currents $\Rightarrow$ repulsion, pushing R away from P (toward Q).

   Force on R due to Q ($d = 0.05~\text{m}$): …

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2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.A long solenoid having 15 cm circumference and 70 turns per metre is carrying a current of 2A. The magnetic field inside the solenoid at a distance 1 cm from the surface and the magnetic field outside the solenoid respectively are (A) $0,\, 1.76 \times 10^{-4}\ \mathrm{T}$ (B) $8.8 \times 10^{-3}\ \mathrm{T},\, 0$ (C) $1.76 \times 10^{-4}\ \mathrm{T},\, 0$ (D) $0,\, 8.8 \times 10^{-3}\ \mathrm{T}$

   ›Reveal solutionSolution

   For an ideal solenoid, the magnetic field is uniform inside and zero outside. The given circumference is irrelevant because the field depends only on turns per metre and current. Inside: $B = \mu_0 n I = 1.76 \times 10^{-4}\ \mathrm{T}$; outside: $B = 0$. The correct option is (C).

   The key concept is the magnetic field of an ideal solenoid. An ideal solenoid is infinitely long and tightly wound, so the field lines are parallel inside and zero outside. The field inside is uniform and given by $B = \mu_0 n I$, where $n$ is the number of turns per unit length and $I$ is the current. The circumference given (15 cm) is a red herring — it does not affect the field magnitude inside or outside for an ideal solenoid.

   1. Identify the relevant parameters.

      • Turns per metre: $n = 70\ \text{turns/m}$
      • Current: $I = 2\ \text{A}$
      • Permeability of free space: $\mu_0 = 4\pi \times 10^{-7}\ \text{T·m/A}$
      • The distance from the surface (1 cm) is irrelevant for an ideal solenoid — inside, the field is uniform everywhere; outside, it is zero everywhere.

   2. Compute the magnetic field inside the solenoid.

      Using $B = \mu_0 n I$:

   $B = (4\pi \times 10^{-7}) \times 70 \times 2$

   First, $70 \times 2 = 140$.

   Then $4\pi \times 10^{-7} \times 140 = 560\pi \times 10^{-7} = 5.6\pi \times 10^{-5}$.

   Using $\pi \approx 3.1416$, $5.6 \times 3.1416 \approx 17.593$, so

   $B \approx 1.7593 \times 10^{-4}\ \mathrm{T} \approx 1.76 \times 10^{-4}\ \mathrm{T}.$

   3. Determine the magnetic field outside the solenoid. …

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3. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.A long horizontal straight wire P carrying a current of 120 A is fixed and another horizontal straight wire Q of linear mass density $1.2 \times 10^{-2}~\mathrm{kg~m^{-1}}$ is placed 2.5 cm below wire P. If the wire Q remains suspended in equilibrium in air, then the current through it is (Acceleration due to gravity $= 10~\mathrm{ms^{-2}}$) (A) 225 A (B) 75 A (C) 250 A (D) 125 A

   ›Reveal solutionSolution

   The key idea is that the magnetic force per unit length on wire Q due to wire P must exactly balance the weight per unit length of wire Q. Using the formula for force between parallel currents, the required current in Q is found to be 125 A.

   The problem is about two parallel current-carrying wires. When two long straight wires carry currents, they exert a magnetic force on each other — attractive if the currents are in the same direction, repulsive if opposite. Here, wire Q is suspended below wire P and remains in equilibrium, meaning the net force on it is zero. The only forces acting on Q are its weight (downward) and the magnetic force from P (which must be upward to balance). So the magnetic force must be repulsive, implying the currents in P and Q are in opposite directions. But the question only asks for the magnitude of the current.

   The magnetic force per unit length between two parallel wires separated by distance $d$, carrying currents $I_1$ and $I_2$, is given by:

   $F_m = \frac{\mu_0 I_1 I_2}{2\pi d}$

   where $\mu_0 = 4\pi \times 10^{-7}~\mathrm{T\,m/A}$.

   For wire Q to be in equilibrium, this upward magnetic force per unit length must equal the downward weight per unit length of Q.

   Let’s work through it step by step.

   1. Identify the given data

      Current in wire P: $I_P = 120~\mathrm{A}$

      Linear mass density of wire Q: $\lambda = 1.2 \times 10^{-2}~\mathrm{kg/m}$

      Separation: $d = 2.5~\mathrm{cm} = 2.5 \times 10^{-2}~\mathrm{m}$

      Acceleration due to gravity: $g = 10~\mathrm{m/s^2}$

      $\mu_0 = 4\pi \times 10^{-7}~\mathrm{T\,m/A}$

   2. Write the equilibrium condition

      Weight per unit length of Q: $w = \lambda g$

      Magnetic force per unit length on Q: $F_m = \frac{\mu_0 I_P I_Q}{2\pi d}$

      For equilibrium: $F_m = w$

      So:

   $\frac{\mu_0 I_P I_Q}{2\pi d} = \lambda g$

   3. Solve for $I_Q$ Rearranging:

   $I_Q = \frac{2\pi d \lambda g}{\mu_0 I_P}$

   Substitute the values:

   $I_Q = \frac{2\pi \times (2.5 \times 10^{-2}) \times (1.2 \times 10^{-2}) \times 10}{(4\pi \times 10^{-7}) \times 120}$ …

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4. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.A solenoid of length 50 cm and radius 10 cm has two closely wound layers of windings 100 turns each. If a current of 2.5 A is passing through the windings, the magnetic field (in $10^{-4}$ T) at a point 5 cm from the axis is (A) $2\pi$ (B) $31.4$ (C) $4\pi$ (D) Zero

   ›Reveal solutionSolution

   The key idea is that inside an ideal solenoid the magnetic field is uniform and axial, independent of radial distance from the axis. For a solenoid with two layers of 100 turns each, total turns = 200, length = 0.5 m, current = 2.5 A. The field at any interior point (including 5 cm from axis) is $B = \mu_0 n I = 4\pi \times 10^{-4}$ T, which matches option (C).

   Concept & Intuition

   The magnetic field inside a long, tightly wound solenoid is nearly uniform and directed along its axis. This is because the field contributions from each turn add constructively inside, while outside they cancel. The standard formula $B = \mu_0 n I$ (where $n$ = turns per unit length) applies for points well inside the solenoid, far from the ends. The radial position (here 5 cm from axis) does not matter as long as it is inside the solenoid — the field is the same everywhere inside. The radius of the solenoid (10 cm) is given only to confirm that the point at 5 cm is indeed inside.

   Step-by-step reasoning

   1. Identify the relevant parameters

      • Length of solenoid: $L = 50\ \text{cm} = 0.5\ \text{m}$
      • Radius: $R = 10\ \text{cm} = 0.1\ \text{m}$
      • Two layers, each with 100 turns → total turns $N = 200$
      • Current: $I = 2.5\ \text{A}$
      • Point of interest: 5 cm from axis (inside the solenoid, since 5 cm < 10 cm)

   2. Compute turns per unit length

   $n = \frac{N}{L} = \frac{200}{0.5} = 400\ \text{turns per meter}$

   3. Apply the formula for the magnetic field inside an ideal solenoid

   $B = \mu_0 n I$

   where $\mu_0 = 4\pi \times 10^{-7}\ \text{T·m/A}$.

   4. Substitute the numbers

   $B = (4\pi \times 10^{-7}) \times 400 \times 2.5$

   First compute $400 \times 2.5 = 1000$. …

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5. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.An alpha particle moving with certain speed towards east enters a uniform magnetic field directed vertically up. The alpha particle will then move in (A) vertical circular path with the same speed (B) horizontal circular path with the same speed (C) vertical circular path with increased speed (D) vertical circular path with decreased speed

   ›Reveal solutionSolution

   The key idea is that the magnetic force is always perpendicular to velocity, so it changes only direction, not speed. For an alpha particle moving east in a vertical magnetic field, the force is horizontal, causing a horizontal circular path at constant speed. The correct option is (B).

   The relevant concept is the magnetic force on a moving charge:

   $\vec{F} = q (\vec{v} \times \vec{B})$

   This force is always perpendicular to both velocity and magnetic field. Because it does no work, the particle’s speed remains constant. The direction of the force determines the plane of the circular motion.

   Let’s work through it step by step:

   1. Identify the directions

      • The alpha particle moves east (say, along the positive x-axis).
      • The magnetic field is vertically up (say, along the positive z-axis).
      • The charge of an alpha particle is positive ($q = +2e$).

   2. Find the direction of the magnetic force

      Use the right-hand rule for $\vec{v} \times \vec{B}$:

      • Point fingers east (velocity), curl them upward (field), thumb points north (positive y-direction).
      • Since the charge is positive, the force is exactly in that direction. So the force is horizontal (north), not vertical.

   3. Determine the motion’s plane

      • The force is perpendicular to velocity, so it acts as a centripetal force.
      • The velocity is east, the force is north — both are horizontal.
      • Therefore, the particle moves in a horizontal circle (in the east–north plane). …

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6. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.An alpha particle moving with certain speed towards east enters a uniform magnetic field directed vertically up. The alpha particle will then move in (A) vertical circular path with decreased speed (B) horizontal circular path with the same speed (C) vertical circular path with increased speed (D) vertical circular path with the same speed

   ›Reveal solutionSolution

   The key is that the magnetic force is always perpendicular to velocity, so it does no work and speed stays constant; using the right-hand rule, the force is horizontal, so the path is a horizontal circle. The correct option is (B).

   The relevant concept is the magnetic force on a moving charge:

   $\vec{F} = q (\vec{v} \times \vec{B})$

   This force is always perpendicular to both the velocity and the magnetic field. Because it’s perpendicular to velocity, it does no work — so the speed of the particle never changes. The force only changes the direction of motion, causing circular motion in the plane perpendicular to the magnetic field.

   1. Identify the directions

      • The alpha particle (positive charge, $q = +2e$) moves east. Let’s call east the $+x$ direction.
      • The magnetic field is directed vertically up. That’s the $+z$ direction (if we take up as positive $z$).

   2. Apply the right-hand rule for magnetic force

      For a positive charge, the force is in the direction of $\vec{v} \times \vec{B}$.

      • $\vec{v}$ points east ($+x$), $\vec{B}$ points up ($+z$).
      • Cross product: $\hat{x} \times \hat{z} = -\hat{y}$ (south). So the force is initially toward the south (horizontal, perpendicular to both east and up).

   3. Determine the plane of motion

      Since the force is always perpendicular to $\vec{B}$, the motion stays in the plane perpendicular to the field. The field is vertical, so the perpendicular plane is horizontal. The particle will move in a horizontal circular path.

   4. Check the speed …

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7. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.A proton moving with a velocity of $8 \times 10^5 \ \mathrm{ms}^{-1}$ enters a uniform magnetic field normal to the direction of the magnetic field. If the radius of the circular path of the proton in the magnetic field is $8.3 \ \mathrm{cm}$, then the magnitude of the magnetic field is (Charge of proton $= 1.6 \times 10^{-19} \ \mathrm{C}$ and mass of the proton $= 1.66 \times 10^{-27} \ \mathrm{kg}$) (A) 500 mT (B) 100 mT (C) 200 mT (D) 400 mT

   ›Reveal solutionSolution

   The magnetic force provides the centripetal force for circular motion, so equating $qvB = \frac{mv^2}{r}$ gives $B = \frac{mv}{qr}$. Substituting the given values yields $B = 0.1\ \mathrm{T} = 100\ \mathrm{mT}$, which corresponds to option (B).

   The key idea is that when a charged particle moves perpendicular to a uniform magnetic field, the magnetic force acts as a centripetal force, causing the particle to move in a circle. The radius of that circle depends on the particle's mass, charge, speed, and the field strength. By equating the two forces, we can solve directly for the magnetic field.

   1. Identify the relevant physics For a charge $q$ moving with speed $v$ perpendicular to a uniform magnetic field $B$, the magnetic force is $F_B = qvB$. This force is always perpendicular to the velocity, so it does no work but instead changes the direction of motion — exactly what is needed for circular motion. The centripetal force required to keep a mass $m$ moving in a circle of radius $r$ is $F_c = \frac{mv^2}{r}$. Setting them equal gives:

   $qvB = \frac{mv^2}{r}$

   Cancel one factor of $v$ (since $v \neq 0$):

   $qB = \frac{mv}{r}$

   So the magnetic field magnitude is:

   $B = \frac{mv}{qr}$

   2. Convert all quantities to SI units

      • Velocity: $v = 8 \times 10^5\ \mathrm{m/s}$ (already SI)
      • Radius: $r = 8.3\ \mathrm{cm} = 8.3 \times 10^{-2}\ \mathrm{m}$
      • Charge: $q = 1.6 \times 10^{-19}\ \mathrm{C}$
      • Mass: $m = 1.66 \times 10^{-27}\ \mathrm{kg}$

   3. Plug in the numbers

   $B = \frac{(1.66 \times 10^{-27})(8 \times 10^5)}{(1.6 \times 10^{-19})(8.3 \times 10^{-2})}$

   First compute the numerator:

   $1.66 \times 8 = 13.28 \quad \Rightarrow \quad 13.28 \times 10^{-22} = 1.328 \times 10^{-21}$

   (since $10^{-27} \times 10^5 = 10^{-22}$)

   Now the denominator:

   $$$$

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8. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.Two charged particles A and B of masses m and 2m, charges 2q and 3q respectively moving with same velocity enter a uniform magnetic field such that both the particles make same angle ($< 90^\circ$) with the direction of the magnetic field. Then the ratio of the pitches of the helical paths of the particles A and B is (A) $4 : 3$ (B) $3 : 2$ (C) $3 : 4$ (D) $2 : 3$

   ›Reveal solutionSolution

   The pitch of a helix in a magnetic field depends only on the parallel velocity component and the cyclotron frequency; since both particles have the same velocity and angle, the pitch ratio equals the inverse ratio of their masses, giving $4:3$.

   The key concept is that when a charged particle enters a uniform magnetic field at an angle, its velocity splits into two components: one parallel to the field (constant) and one perpendicular (causing circular motion). The pitch is the distance traveled along the field direction in one full cyclotron period. Since the parallel velocity is the same for both particles (same speed and same angle), the pitch ratio reduces to the ratio of their cyclotron periods, which is simply the ratio of their masses.

   1. Write the pitch formula. The pitch $P$ of a helical path is given by

   $P = v_{\parallel} \cdot T$

   where $v_{\parallel} = v \cos\theta$ is the velocity component parallel to the magnetic field, and $T$ is the cyclotron period.

   2. Express the cyclotron period. The cyclotron period for a particle of mass $m$ and charge $q$ in a magnetic field $B$ is

   $T = \frac{2\pi m}{qB}.$

   This comes from the angular frequency $\omega = \frac{qB}{m}$.

   3. Combine to get pitch in terms of given quantities.

   $P = (v \cos\theta) \cdot \frac{2\pi m}{qB}.$

   Notice that $v$, $\theta$, and $B$ are the same for both particles.

   4. Find the ratio for particles A and B. For particle A: $m_A = m$, $q_A = 2q$ → $P_A = \frac{2\pi m v \cos\theta}{2q B} = \frac{\pi m v \cos\theta}{q B}.$ …

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9. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.A straight wire carrying a current of $2\sqrt{2}$ A is making an angle of $45^\circ$ with the direction of uniform magnetic field of $3$ T. The force per unit length on the wire due to the magnetic field is (A) $4~\mathrm{Nm}^{-1}$ (B) $8~\mathrm{Nm}^{-1}$ (C) $6~\mathrm{Nm}^{-1}$ (D) $3~\mathrm{Nm}^{-1}$

   ›Reveal solutionSolution

   The force per unit length on a current-carrying wire in a magnetic field is given by $F/L = I B \sin\theta$. Substituting $I = 2\sqrt{2}\,\text{A}$, $B = 3\,\text{T}$, and $\theta = 45^\circ$ gives $F/L = 6\,\text{N/m}$, so the correct option is (C).

   The key concept here is the magnetic force on a current-carrying wire in a uniform field. The force depends not only on the current and field strength but also on the angle between the wire and the field — only the component of the current perpendicular to the field experiences the force. This is captured by the cross product $\vec{F} = I \vec{L} \times \vec{B}$, whose magnitude is $I L B \sin\theta$.

   Let’s work through it step by step.

   1. Recall the formula for magnetic force on a straight wire. For a wire of length $L$ carrying current $I$ in a uniform magnetic field $B$, the magnitude of the force is

   $F = I L B \sin\theta$

   where $\theta$ is the angle between the direction of the current and the magnetic field.

   The force per unit length is then

   $\frac{F}{L} = I B \sin\theta.$

   2. Identify the given values.

      • Current, $I = 2\sqrt{2}$ A
      • Magnetic field, $B = 3$ T
      • Angle, $\theta = 45^\circ$

   3. Compute $\sin 45^\circ$.

   $\sin 45^\circ = \frac{\sqrt{2}}{2}$

   4. Plug into the formula.

   $\frac{F}{L} = (2\sqrt{2}) \times 3 \times \frac{\sqrt{2}}{2}$

   5. Simplify step by step. Multiply the numerical parts: $$$$

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10. TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
    Q.Two long straight parallel conductors are 2 cm apart and carry currents of 5 A and 10 A in opposite directions. The force per unit length of each conductor is (A) $5 \times 10^{-6} \ \mathrm{Nm}^{-1}$ (B) $5 \times 10^{-4} \ \mathrm{Nm}^{-1}$ (C) $2 \times 10^{-6} \ \mathrm{Nm}^{-1}$ (D) $5 \times 10^{-8} \ \mathrm{Nm}^{-1}$

    ›Reveal solutionSolution

    When two parallel conductors carry currents in opposite directions, they repel each other. The force per unit length is calculated using the formula $F/L = \mu_0 I_1 I_2 / (2\pi d)$, which gives a value of $\boxed{5 \times 10^{-4} \ \mathrm{Nm}^{-1}}$.

    The fundamental concept here is that a current-carrying wire produces a magnetic field around it. When another current-carrying wire is placed within this magnetic field, it experiences a magnetic force. This is a direct consequence of the Lorentz force law.

    For two long, straight, parallel conductors:

    1. One conductor (say, wire 1) carrying current $I_1$ creates a magnetic field in the space around it. The magnetic field lines form concentric circles around the wire.
    2. The second conductor (wire 2) carrying current $I_2$ is situated in this magnetic field. Since wire 2 is carrying a current, its moving charges interact with the magnetic field produced by wire 1, resulting in a force on wire 2.
    3. Crucially, the force on wire 1 due to the magnetic field of wire 2 will be equal in magnitude and opposite in direction to the force on wire 2 due to the magnetic field of wire 1, as per Newton's third law.

    The direction of the force depends on the relative direction of the currents:

    • If the currents are in the same direction, the wires attract each other.
    • If the currents are in opposite directions, the wires repel each other. In this problem, the currents are in opposite directions, so the conductors will repel each other. The question asks for the magnitude of the force per unit length.

    The magnitude of the force per unit length ($F/L$) between two long, straight, parallel conductors carrying currents $I_1$ and $I_2$ separated by a distance $d$ is given by:

    $\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$

    where $\mu_0$ is the permeability of free space, with a value of $4\pi \times 10^{-7} \ \mathrm{Tm/A}$.

    ›Proof

    Derivation of the formula

    1. Consider a long straight wire (wire 1) carrying current $I_1$. The magnetic field $B_1$ produced by this wire at a perpendicular distance $d$ from it is given by Ampere's law: $B_1 = \frac{\mu_0 I_1}{2\pi d}$ The direction of $B_1$ can be found using the right-hand thumb rule.
    2. Now, consider a second long straight wire (wire 2) carrying current $I_2$, placed parallel to wire 1 at a distance $d$. Wire 2 is immersed in the magnetic field $B_1$ produced by wire 1.
    3. The force $F_2$ experienced by a length $L$ of wire 2 due to the magnetic field $B_1$ is given by the Lorentz force law for a current-carrying conductor: $\vec{F}_2 = I_2 (\vec{L} \times \vec{B}_1)$ Since wire 2 is parallel to wire 1, and the magnetic field $B_1$ circles wire 1, the current $\vec{L}$ in wire 2 is perpendicular to the magnetic field $\vec{B}_1$ at the location of wire 2. Therefore, the angle between $\vec{L}$ and $\vec{B}_1$ is $90^\circ$, and $\sin 90^\circ = 1$.
    4. The magnitude of the force on length $L$ of wire 2 is: $F_2 = I_2 L B_1$
    5. Substitute the expression for $B_1$: $F_2 = I_2 L \left( \frac{\mu_0 I_1}{2\pi d} \right)$
    6. The force per unit length ($F/L$) on wire 2 is then: $\frac{F_2}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$ …

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11. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
    Q.A current carrying loop is placed in a uniform magnetic field ‘B’ in different orientations I, II, III and IV as shown in the figure. The correct order of decreasing potential energy is ($\hat{n}$- unit vector normal to the plane of the loop) (A) I, III, II, IV (B) I, II, III, IV (C) I, IV, II, III (D) III, IV, I, II

    ›Reveal solutionSolution

    Use $U=-mB\cos\theta$: energy falls monotonically as the angle between $\hat n$ and $\vec B$ shrinks from $180^\circ$ to $0^\circ$. Ranking the four pictures by that angle gives I, IV, II, III — option (C).

    The concept first

    A current loop of area $A$ carrying current $I$ behaves like a tiny bar magnet with magnetic moment

    $\vec m = IA\,\hat n ,$

    where $\hat n$ is the unit normal fixed by the right-hand rule from the current sense. In an external field the loop stores orientational potential energy

    $U=-\vec m\cdot\vec B=-mB\cos\theta .$

    Why the minus sign? Because a magnet wants to line up with the field. Aligned ($\theta=0$) is the state of lowest energy, $U=-mB$ — stable equilibrium. Anti-aligned ($\theta=180^\circ$) is the state of highest energy, $U=+mB$ — unstable equilibrium (nudge it and it flips). Perpendicular is exactly halfway: $U=0$, and this is where the torque $\tau=mB\sin\theta$ is largest.

    The key monotonic fact: as $\theta$ decreases from $180^\circ$ to $0^\circ$, $\cos\theta$ increases, so $U=-mB\cos\theta$ decreases steadily. Ranking energy is therefore just ranking the angle $\theta$, from biggest to smallest.

    Step-by-step

    1. Orientation I. $\hat n$ points left, $\vec B$ points right — they lie on the same line, opposite senses: $\theta=180^\circ$.

    $U_{\text{I}}=-mB\cos180^{\circ}=+mB\quad(\text{maximum possible}).$

    2. Orientation IV. $\hat n$ points up-and-to-the-left while $\vec B$ points right — an obtuse angle (between $90^\circ$ and $180^\circ$). So $\cos\theta<0$ and

    $0<U_{\text{IV}}<mB .$

    It sits below I but above the perpendicular case. …

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12. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
    Q.Assertion (A) : The magnetic field lines are continuous and form closed loops. Reason (R) : Magnetic monopole does not exist. The correct option among the following is (A) (A) is true, (R) is true and (R) is the correct explanation for (A) (B) (A) is true, (R) is true but (R) is not the correct explanation for (A) (C) (A) is true but (R) is false (D) (A) is false but (R) is true

    ›Reveal solutionSolution

    Both statements are true, and the non-existence of magnetic monopoles is the direct cause of the closed-loop nature of magnetic field lines (Gauss's law for magnetism). Option (A).

    The concept first

    Compare the two fields:

    • Electric field. Charges come singly. A field line starts on a positive charge and ends on a negative one, so Gauss's law reads $\oint \vec E \cdot d\vec S = \dfrac{q_{\text{enc}}}{\varepsilon_0}$ — a net flux exists whenever a charge is enclosed.
    • Magnetic field. Despite long searches, an isolated magnetic pole has never been observed. Cut a bar magnet in half and each piece is again a complete dipole with its own N and S. Consequently Gauss's law for magnetism is

    $\oint_S \vec{B}\cdot d\vec{S} = 0 \qquad \text{(equivalently } \nabla\cdot\vec{B} = 0\text{)}$

    for every closed surface.

    Step-by-step

    1. Is the Reason true? Yes — no magnetic monopole has ever been detected; magnetic sources are always dipoles (current loops or spins). (R) is true.
    2. Is the Assertion true? Yes. Whatever magnetic field lines you draw — around a bar magnet, a solenoid, a straight current-carrying wire — they never terminate. Outside a bar magnet they run N → S; inside the magnet they continue S → N, completing the loop. Around a wire they are perfect circles. (A) is true. …

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