Q.Consider a wire carrying a steady current, $I$ placed in a uniform magnetic field $\vec{B}$ perpendicular to its length. Consider the charges inside the wire. It is known that magnetic forces do no work. This implies that,

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1. COMEDK 2026Set 2026-A1 markMCQ
   Q.A circular coil of radius $r=10 c m$ having 300 turns carries a current of 2 A . The coil is suspended vertically in a uniform magnetic field of strength 0.7 T . If the plane of the coil makes an angle $30^{\circ}$ with the magnetic field, the torque needed to prevent it from turning is: (A) 11.42 Nm (B) 1.1 Nm (C) 22.84 Nm (D) 5.71 Nm

   ›Reveal solutionSolution

   The torque on a current-carrying coil in a magnetic field is given by $\tau = N I A B \sin\theta$, where $\theta$ is the angle between the normal to the coil and the field. Here the plane makes $30^\circ$ with the field, so the normal makes $60^\circ$, giving $\tau \approx 11.42\ \text{Nm}$. The correct option is (A).

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   Concept & Intuition

   A current loop in a magnetic field experiences a torque that tries to align its normal (the perpendicular to its plane) with the field. The formula is $\tau = N I A B \sin\phi$, where $\phi$ is the angle between the normal and $\vec{B}$.

   The common pitfall: the problem gives the angle between the plane of the coil and the field, not the normal. If you use $30^\circ$ directly, you get the wrong answer. Always convert: $\phi = 90^\circ - \text{(plane angle)}$.

   ---

   Step-by-step solution

   1. Identify given quantities

      • Radius $r = 10\ \text{cm} = 0.1\ \text{m}$
      • Number of turns $N = 300$
      • Current $I = 2\ \text{A}$
      • Magnetic field $B = 0.7\ \text{T}$
      • Angle between plane of coil and $\vec{B}$: $\theta_{\text{plane}} = 30^\circ$

   2. Find the area of the coil

   $A = \pi r^2 = \pi (0.1)^2 = 0.01\pi\ \text{m}^2$

   3. Determine the correct angle $\phi$ The torque formula uses the angle between the normal to the coil and $\vec{B}$. Since the normal is perpendicular to the plane:

   $\phi = 90^\circ - \theta_{\text{plane}} = 90^\circ - 30^\circ = 60^\circ$

   4. Apply the torque formula

   $\tau = N I A B \sin\phi$

   Substitute values:

   $\tau = 300 \times 2 \times (0.01\pi) \times 0.7 \times \sin 60^\circ$

   $\tau = 300 \times 2 \times 0.01\pi \times 0.7 \times \frac{\sqrt{3}}{2}$

   5. Simplify step by step
      • $300 \times 2 = 600$
      • $600 \times 0.01 = 6$ …

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.Two long parallel conductors $X$ and $Y$ are placed vertically at a distance $d$ apart. Conductor X carries a current I upwards and conductor Y carries a current 21 downwards. A third long conductor Z is placed parallel to both X and Y and between X and Y . If Z carries a current I upwards and is at a distance $x$ from conductor X , then: A. All the conductors are in equilibrium and the net force on Z is zero B. The net force on Z is $\frac{\mu_0 I^2}{2 x \pi}\left[\frac{d+x}{d-x}\right]$ towards X C. The net force on Z is $\frac{\mu_0 I^2}{2 x \pi}\left[\frac{d+x}{d-x}\right]$ towards Y D. The net force on Z is $\frac{\mu_0 I^2}{2 x \pi}\left[\frac{d-x}{d+x}\right]$ towards Y (A) B (B) D (C) A (D) C

   ›Reveal solutionSolution

   Z (current $I$ up) is attracted by X (current $I$ up) and repelled by Y (current $2I$ down); both pushes point toward X, so they add. The net force is $\dfrac{\mu_0 I^2}{2\pi x}\!\left[\dfrac{d+x}{d-x}\right]$ towards X — that is statement B, which is choice (A).

   Concept. Force per unit length between long parallel currents is $F=\dfrac{\mu_0 I_1 I_2}{2\pi r}$: attractive for parallel currents, repulsive for antiparallel.

   Step 1 — Force on Z from X. Both up (parallel) $\Rightarrow$ attractive; separation $x$:

   $F_{ZX}=\frac{\mu_0 I\cdot I}{2\pi x}=\frac{\mu_0 I^2}{2\pi x}\quad(\text{toward X}).$

   Step 2 — Force on Z from Y. Z up, Y down (antiparallel) $\Rightarrow$ repulsive; Y is on the far side at distance $d-x$, so Z is pushed away from Y, i.e. also toward X:

   $F_{ZY}=\frac{\mu_0 I\cdot 2I}{2\pi (d-x)}=\frac{\mu_0 I^2}{\pi (d-x)}\quad(\text{toward X}).$

   Step 3 — Both forces point toward X, so they add. …

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3. KCET 2025Set D-41 markMCQ
   Q.A square loop of side $2\,\mathrm{m}$ lies in the Y-Z plane in a region having a magnetic field $\vec{B} = (5\hat{i} + 3\hat{j} - 4\hat{k})\,\mathrm{T}$. The magnitude of magnetic flux through the square loop is (A) $20\,\mathrm{Wb}$ (B) $12\,\mathrm{Wb}$ (C) $16\,\mathrm{Wb}$ (D) $10\,\mathrm{Wb}$

   ›Reveal solutionSolution

   Flux is the dot product $\vec{B}\cdot\vec{A}$; the area vector of a Y–Z-plane loop is along $\hat i$, so only $B_x$ survives.

   Step 1 — The concept: magnetic flux is a dot product.

   For a uniform field over a plane loop,

   $\Phi_B = \vec{B}\cdot\vec{A} = BA\cos\theta$

   where $\vec{A}$ is the area vector — magnitude equal to the area, direction along the normal to the plane of the loop. The dot product is what picks out only the component of $\vec B$ that pierces the loop; a component lying in the plane of the loop threads nothing through it.

   Step 2 — Find the area vector.

   The loop lies in the Y–Z plane. The normal to the Y–Z plane is the x-direction, so

   $\hat{n} = \hat{i}$

   The side is $2\,\mathrm{m}$, so the area is

   $A = (2\,\mathrm{m})^2 = 4\,\mathrm{m^2} \quad\Longrightarrow\quad \vec{A} = 4\hat{i}\ \mathrm{m^2}$

   Step 3 — Take the dot product.

   $\vec{B} = (5\hat{i} + 3\hat{j} - 4\hat{k})\ \mathrm{T}$

   $\Phi_B = \vec{B}\cdot\vec{A} = (5\hat{i} + 3\hat{j} - 4\hat{k})\cdot(4\hat{i})$

   Using $\hat i\cdot\hat i = 1$ and $\hat j\cdot\hat i = \hat k\cdot\hat i = 0$:

   $\Phi_B = 5 \times 4 = 20\ \mathrm{Wb}$

   Step 4 — Sanity check on the discarded components. …

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4. COMEDK 2025Set 2025-A1 markMCQ
   Q.A rectangular coil of length 10 cm and breadth 9 cm carries a current of 10 A . A long straight conductor carrying a current of 20 A is placed 1 cm from the coil parallel to its length and in the same plane of the coil. What will be the net force acting on the straight conductor? (A) $1.8 \times 10^{-4} \mathrm{~N}$ (B) $3.6 \times 10^{-4} \mathrm{~N}$ (C) $1.8 \times 10^{-5} \mathrm{~N}$ (D) $3.6 \times 10^{-5} \mathrm{~N}$

   ›Reveal solutionSolution

   The net force on the straight conductor is found by summing the magnetic forces from each side of the rectangular coil; due to symmetry, the forces from the two sides parallel to the conductor cancel, leaving only the force from the nearer perpendicular side, giving $3.6 \times 10^{-4} \, \mathrm{N}$.

   Concept & Intuition

   The problem involves two current-carrying conductors: a long straight wire and a rectangular coil. The force on a current-carrying conductor in a magnetic field is given by $\vec{F} = I \, \vec{L} \times \vec{B}$. Here, the magnetic field at the straight conductor is produced by the rectangular coil. Since the coil has four sides, each side contributes a magnetic field at the location of the straight wire. However, because the straight wire is parallel to the coil’s length and placed 1 cm away, the forces from the two long sides of the coil (parallel to the straight wire) will be equal in magnitude but opposite in direction — they cancel. The net force thus comes only from the two short sides (breadth) of the coil, but one is much farther away, so its contribution is negligible compared to the nearer one. We therefore compute the force due to the nearer breadth side.

   Step-by-step solution

   1. Set up the geometry and currents

      Coil: length $l = 10 \, \text{cm} = 0.1 \, \text{m}$, breadth $b = 9 \, \text{cm} = 0.09 \, \text{m}$, current $I_c = 10 \, \text{A}$.

      Straight conductor: current $I_s = 20 \, \text{A}$, placed parallel to the coil’s length at a distance $d = 1 \, \text{cm} = 0.01 \, \text{m}$ from the nearer side of the coil, in the same plane.

   2. Identify which sides of the coil produce a net force

      The straight conductor is parallel to the two long sides (length $l$). The magnetic field from a long straight wire at a distance $r$ is $B = \frac{\mu_0 I}{2\pi r}$, directed tangentially. For the two long sides, the fields at the straight wire are equal in magnitude but opposite in direction (one attracts, one repels), so they cancel exactly.

      The two short sides (breadth $b$) are perpendicular to the straight wire. The nearer short side is at distance $d = 0.01 \, \text{m}$; the farther short side is at distance $d + l = 0.11 \, \text{m}$. The force from the farther side is much smaller (by a factor of 11) and can be neglected for the net force calculation.

   3. Compute the magnetic field at the straight wire due to the nearer short side

      The short side is a straight wire of length $b = 0.09 \, \text{m}$ carrying current $I_c = 10 \, \text{A}$. The straight conductor is at a perpendicular distance $d = 0.01 \, \text{m}$ from the midpoint of this side? Actually, careful: The straight wire is placed 1 cm from the coil, parallel to the length. That means it runs alongside the coil, 1 cm away from the nearer long side. The short side is perpendicular to the straight wire, so the distance from the straight wire to the short side varies along the short side. However, since the straight wire is long and the short side is short, we can approximate the short side as a finite straight segment. The magnetic field at a point opposite the center of a finite wire of length $b$ at perpendicular distance $d$ is

   $B = \frac{\mu_0 I}{4\pi d} \left( \sin\theta_1 + \sin\theta_2 \right)$

   where $\theta_1$ and $\theta_2$ are the angles from the point to the ends of the wire. Here, the straight wire is parallel to the coil’s length, so it lies along the direction of the long side. The short side is perpendicular. The point on the straight wire closest to the short side is at the midpoint of the short side? Actually, the straight wire runs the entire length of the coil (10 cm) and beyond, so it is opposite the entire short side. The distance from the straight wire to the short side is constant = 1 cm? No — the straight wire is 1 cm from the long side, so the distance to the short side is also 1 cm at the corner, but varies. However, because the straight wire is long and the short side is short, the dominant contribution is from the segment of the straight wire directly opposite the short side. For simplicity, treat the short side as a long straight wire (since $b \gg d$? Actually $b=9$ cm, $d=1$ cm, so $b$ is 9 times $d$, so it's reasonably long). Use the formula for an infinite straight wire as an approximation:

   $B \approx \frac{\mu_0 I_c}{2\pi d}$

   This is a good approximation because the short side is much longer than the distance.

   So

   $$$$

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5. COMEDK 2025Set 2025-E1 markMCQ
   Q.A current carrying closed loop in the form of a right isosceles triangle XYZ is placed in a uniform magnetic field $B$ acting along $X Y$ of the loop. If the magnetic force on the arm $Y Z$ is $\sqrt{2} F$, then the force on the arm $X Z$ is: (A) $-\sqrt{2} F$ (B) $\frac{\mathrm{F}}{\sqrt{2}}$ (C) $+2 \sqrt{2} F$ (D) $-F$

   ›Reveal solutionSolution

   The net magnetic force on a current-carrying closed loop in a uniform field is always zero. Using the given force on YZ and the geometry of the right isosceles triangle, the force on XZ must be the negative of the vector sum of the forces on the other two sides, giving the answer as $-\sqrt{2}F$, which is option (A).

   ---

   Concept and Intuition

   The key idea is a beautiful and powerful result: In a uniform magnetic field, the total magnetic force on any closed current-carrying loop is exactly zero. Why? Because the force on each infinitesimal current element is $\vec{F} = I \, d\vec{\ell} \times \vec{B}$, and when you integrate $d\vec{\ell}$ around a closed loop, you get the zero vector. So the vector sum of the forces on all three sides of the triangle must vanish:

   $\vec{F}_{XY} + \vec{F}_{YZ} + \vec{F}_{XZ} = 0$

   We are told the force on YZ is $\sqrt{2}F$ (presumably in some direction), and we need the force on XZ. If we can find the force on XY, we can solve for XZ. But the magnetic field is along XY — that gives us a huge shortcut.

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   Step-by-step reasoning

   1. Identify the magnetic field direction.

      The field $\vec{B}$ acts along XY. In the given figure, XY is vertical (from X at top-left to Y at bottom-left). So $\vec{B}$ points vertically downward (or upward — the exact sign doesn't matter for magnitude, but we must be consistent with vector directions).

   2. Force on side XY.

      The current in side XY flows along the same line as $\vec{B}$ (both along XY). The magnetic force on a current element is $\vec{F} = I \vec{L} \times \vec{B}$. If $\vec{L}$ and $\vec{B}$ are parallel (or antiparallel), their cross product is zero.

      Therefore, the force on arm XY is zero: $\vec{F}_{XY} = 0$.

   3. Force on side YZ.

      YZ is horizontal (from Y to Z). The field $\vec{B}$ is vertical. So $\vec{L}_{YZ}$ is perpendicular to $\vec{B}$. The magnitude of the force is $F_{YZ} = I L_{YZ} B \sin 90^\circ = I L_{YZ} B$.

      The problem states this force is $\sqrt{2}F$. So we have:

   $I L_{YZ} B = \sqrt{2}F$

   The direction of this force is given by the right-hand rule: if current flows from Y to Z (say to the right) and $\vec{B}$ is downward, then $\vec{F}_{YZ}$ points out of the page (or into it, depending on sign conventions). We don't need the exact direction for the magnitude, only that it is some vector we'll call $\vec{F}_{YZ}$.

   4. Force on side XZ (the hypotenuse). Since the net force on the loop is zero: $$$$

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6. KCET 2024Set D-21 markMCQ
   Q.A particle of specific charge $\dfrac{q}{m} = \pi \mathrm{Ckg}^{-1}$ is projected from the origin towards positive x-axis with the velocity $10~\mathrm{ms}^{-1}$ in a uniform magnetic field $\vec{B} = -2\hat{k}~\mathrm{T}$. The velocity $\vec{v}$ particle after time $t = \dfrac{1}{12}~\mathrm{s}$ will be (in ms$^{-1}$) (A) $5\left(\hat{i}+\hat{j}\right)$ (B) $5\left(\hat{i}+\sqrt{3}\hat{j}\right)$ (C) $5\left(\sqrt{3}\hat{i}-\hat{j}\right)$ (D) $5\left(\sqrt{3}\hat{i}+\hat{j}\right)$

   ›Reveal solutionSolution

   A magnetic force does no work, so the speed stays $10\ \text{ms}^{-1}$ and the velocity merely rotates through $\theta = \omega t$ with $\omega = qB/m$; find the sense of rotation from $q\vec v\times\vec B$.

   Step 1 — The concept: uniform circular motion

   Since $\vec{F} = q\,\vec{v}\times\vec{B}$ is always perpendicular to $\vec v$, it does no work: the speed is constant and the velocity vector just rotates uniformly. Here $\vec v \perp \vec B$ (velocity along $x$, field along $-z$), so the path is a full circle with angular frequency

   $\omega = \frac{qB}{m} = \left(\frac{q}{m}\right)B$

   Step 2 — Compute the angular frequency

   $\frac{q}{m} = \pi\ \text{C kg}^{-1}, \qquad |\vec{B}| = 2\ \text{T}$

   $\omega = \pi \times 2 = 2\pi\ \text{rad s}^{-1}$

   (So the period is $T = 2\pi/\omega = 1\ \text{s}$ — a neat check.)

   Step 3 — Angle turned in the given time

   $\theta = \omega t = (2\pi)\left(\frac{1}{12}\right) = \frac{\pi}{6} = 30^\circ$

   (Equivalently: one-twelfth of a $1\ \text{s}$ period is one-twelfth of a full circle $= 30^\circ$.)

   Step 4 — Which way does it turn?

   Initially $\vec{v} = 10\,\hat{i}$ and $\vec{B} = -2\hat{k}$. The magnetic force (taking $q>0$, as the positive specific charge implies) is

   $\vec{F} = q\,\vec{v}\times\vec{B} = q\left(10\hat{i}\right)\times\left(-2\hat{k}\right) = -20q\,(\hat{i}\times\hat{k})$

   Since $\hat{i}\times\hat{k} = -\hat{j}$:

   $\vec{F} = -20q(-\hat{j}) = +20q\,\hat{j}$

   The force is along $+\hat{j}$, so the velocity is pulled from $+x$ towards $+y$ — an anticlockwise rotation in the x–y plane.

   Step 5 — Write the velocity after $30^\circ$ …

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7. COMEDK 2024Set 2024-E1 markMCQ
   Q.A wire of length $2 \mathrm{~m}$ carries a current of $1 \mathrm{~A}$ along the $\mathrm{x}$ axis. A magnetic field $B=B_0(i+j+k)$ tesla exists in space. The magnitude of magnetic force on the wire is (A) 2B$_0$ N (B) Zero (C) 3B$_0$ N (D) 2$\sqrt2$B$_0$ N

   ›Reveal solutionSolution

   The magnetic force on a current-carrying wire is given by $\vec{F} = I \vec{L} \times \vec{B}$. Here the wire lies along the x‑axis, so only the y‑ and z‑components of $\vec{B}$ contribute. The magnitude works out to $2\sqrt{2} B_0$ N, which corresponds to option (D).

   Concept & Intuition

   The force on a straight current-carrying wire in a uniform magnetic field is the cross product of the current direction (vector $\vec{L}$) and the field $\vec{B}$. Only the part of $\vec{B}$ perpendicular to the wire produces a force. Since the wire is along the x‑axis, the x‑component of $\vec{B}$ is parallel to the wire and contributes nothing. The y‑ and z‑components are perpendicular, so they combine to give a net force.

   Step‑by‑step solution

   1. Identify the given quantities

      • Wire length: $L = 2\ \text{m}$ along the x‑axis → $\vec{L} = 2\ \hat{i}\ \text{m}$
      • Current: $I = 1\ \text{A}$
      • Magnetic field: $\vec{B} = B_0 (\hat{i} + \hat{j} + \hat{k})\ \text{T}$

   2. Write the magnetic force formula

      For a straight wire in a uniform field:

   $\vec{F} = I (\vec{L} \times \vec{B})$

   3. Compute the cross product

   $\vec{L} \times \vec{B} = (2\hat{i}) \times (B_0\hat{i} + B_0\hat{j} + B_0\hat{k})$

   Using $\hat{i}\times\hat{i}=0$, $\hat{i}\times\hat{j}=\hat{k}$, $\hat{i}\times\hat{k}=-\hat{j}$:

   $\vec{L} \times \vec{B} = 2B_0 (0 + \hat{k} - \hat{j}) = 2B_0 (-\hat{j} + \hat{k})$

   4. Multiply by current $$$$

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8. COMEDK 2023Set 2023-E1 markMCQ
   Q.A circular loop of area $0.04 \mathrm{~m}^2$ carrying a current of $10 \mathrm{~A}$ is held with its plane perpendicular to a magnetic field induction $0.4 \mathrm{~T}$ Then the torque acting on the circular loop is : (A) 0.004 Nm (B) zero (C) 0.04 Nm (D) 0.02 Nm

   ›Reveal solutionSolution

   The magnetic moment is along the loop's normal; with the plane $\perp B$ the moment is parallel to $B$, so $\tau=IAB\sin0=0$.

   The torque on a current-carrying loop in a magnetic field is

   $\tau = \vec{m}\times\vec{B} = IAB\sin\theta$

   where $\theta$ is the angle between the magnetic dipole moment $\vec m$ (directed along the normal to the loop's plane) and $\vec B$. …

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9. COMEDK 2023Set 2023-M1 markMCQ
   Q.A straight wire of length $2 \mathrm{~m}$ carries a current of $10 \mathrm{~A}$. If this wire is placed in uniform magnetic field of $0.15 \mathrm{~T}$ making an angle of $45^{\circ}$ with the magnetic field, the applied force on the wire will be (A) $1.5 \mathrm{~N}$ (B) $3 \mathrm{~N}$ (C) $3 \sqrt{2} \mathrm{~N}$ (D) $3 / \sqrt{2} \mathrm{~N}$

   ›Reveal solutionSolution

   The magnetic force on the wire is $F=BIL\sin\theta=3/\sqrt2$ N.

   The force on a current-carrying straight wire is

   $F=BIL\sin\theta=(0.15)(10)(2)\sin45^\circ.$ …

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10. COMEDK 2022Set 20221 markMCQ
    Q.A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5 A, then the average force on each electron in the coil due to the magnetic field is (A) $2.5\times10^{-25}$ N (B) $4.5\times10^{-25}$ N (C) $5\times10^{-25}$ N (D) $5.5\times10^{-25}$ N

    ›Reveal solutionSolution

    (Note the number of turns and the radius are not needed for this part - they only matter for the total torque/force, both of which are zero.)

    Concept (NCERT problem): the magnetic force on a single electron is F = e v_d B, where v_d is the drift speed, obtained from I = n e A v_d. (Standard data used with this problem: n = 10^29 m^-3, wire cross-section A = 10^-5 m^2.)

    Drift speed:

    v_d = I / (n e A) = 5 / (10^29 x 1.6 x 10^-19 x 10^-5)

    = 5 / (1.6 x 10^5) = 3.125 x 10^-5 m/s

    Average force on each electron (velocity perpendicular to B): …

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11. KCET 2021Set B-21 markMCQ
    Q.A strong magnetic field is applied on a stationary electron. Then the electron (A) Moves in the direction of the field (B) Moves in an opposite direction of the field (C) Remains stationary (D) Starts spinning

    ›Reveal solutionSolution

    The magnetic force $\vec{F} = q(\vec{v}\times\vec{B})$ is proportional to velocity; a charge at rest feels no magnetic force, however strong the field.

    1. The concept — the Lorentz magnetic force.

    A charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ experiences

    $\vec{F} = q\,(\vec{v}\times\vec{B}), \qquad |F| = qvB\sin\theta$

    The crucial feature — and the whole point of the question — is that $\vec{v}$ appears inside the cross product. A magnetic field acts only on moving charges; it is fundamentally different from an electric field, which acts on a charge whether it moves or not ($\vec{F} = q\vec{E}$).

    2. Apply it to a stationary electron.

    $\vec{v} = 0 \quad\Longrightarrow\quad \vec{F} = q\,(0 \times \vec{B}) = 0$

    The field strength $B$ multiplies zero — so however "strong" the field, the force is exactly zero.

    3. Consequence.

    By Newton's second law, $\vec{a} = \vec{F}/m = 0$. With no acceleration and zero initial velocity, the electron simply remains at rest. …

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12. COMEDK 2021Set 2021-B1 markMCQ
    Q.A wire carrying a current I is placed in the form of a curve $y = \alpha\,\sin\left(\dfrac{\pi x}{L}\right)$, $0 \le x \le 2L$. In the magnetic field of strength B. The force acting on the wire is [FIGURE: a sine curve $y = \alpha\sin(\pi x/L)$ plotted over $0 \le x \le 2L$, showing one full period (positive hump then negative hump) in a region of magnetic field B directed out of the plane (dots)] (A) IBL (B) $\dfrac{IBL}{\pi}$ (C) $IBL\pi$ (D) 2IBL

    ›Reveal solutionSolution

    In a uniform field, $\vec{F} = I\vec{L}_{\text{eff}}\times\vec{B}$ where $\vec{L}_{\text{eff}}$ is the straight line from start to end. Here that length is $2L$, so $F = 2IBL$.

    For any shaped wire in a uniform magnetic field, the net force equals the force on the straight segment connecting its endpoints:

    $\vec{F} = I\,\vec{L}_{\text{eff}} \times \vec{B}.$ …

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