Q.A particle of charge q moves with a velocity v at angle θ to a magnetic field B. What is the force experienced by the particle? OR What is the unit of magnetic pole-strength?

Showing the 12 most recent of 38 on this concept.

1. CBSE 2026Set ANNUAL1 mark
   Q.Calculate the magnitude of force experienced by a stationary charge exposed to uniform magnetic field.

   ›Reveal solutionSolution

   The magnetic Lorentz force is $F = qvB\sin\theta$; for a stationary charge $v=0$, so the force is exactly zero.

   The force experienced by a charge $q$ moving with velocity $v$ in a magnetic field $B$ is:

   $\vec F = q\vec v \times \vec B, \qquad |F| = qvB\sin\theta$

   For a stationary charge, $v = 0$. Substituting:

   $F = q(0)B\sin\theta = 0$

   …

   🔒 Start your 14-day free trial to see the full solution →
2. CBSE 2025Set ANNUAL1 markMCQ
   Q.'Tesla' is unit of which of the following physical quantity-

   (a) Electric Field

   (b) Electric Dipole moment

   (c) Magnetic Field

   (d) Magnetic Moment

   ›Reveal solutionSolution

   The tesla (T) is the SI unit of magnetic field (magnetic flux density).

   The tesla is defined from the magnetic force law $F = qvB\sin\theta$: $1\text{ T} = 1\ \text{N}\,\text{A}^{-1}\,\text{m}^{-1} = 1\ \text{Wb/m}^2$. It measures magnetic field strength $B$. Electric …

   🔒 Start your 14-day free trial to see the full solution →
3. CBSE 2025Set ANNUAL1 markMCQ
   Q.An electric charge q is moving in a uniform magnetic field B parallel to the lines of force with velocity v. The magnetic force acting on the charge is

   (a) Zero

   (b) qvB

   (c) qB/v

   (d) qv/B

   ›Reveal solutionSolution

   A charge moving exactly along the magnetic field direction experiences no magnetic force, because the force depends on $\sin\theta$ between $v$ and $B$.

   The magnetic (Lorentz) force on a moving charge is:

   $\vec{F} = q\,\vec{v}\times\vec{B}$, with magnitude $F = qvB\sin\theta$

   …

   🔒 Start your 14-day free trial to see the full solution →
4. CBSE 2025Set ANNUAL1 mark
   Q.At what angle should a proton move in a uniform magnetic field so that the proton continues to move in its initial direction?

   ›Reveal solutionSolution

   The magnetic force on a moving charge is F = qv x B, whose magnitude is qvB sin(theta); this is zero only when theta = 0 degrees or 180 degrees, i.e. v is along (or exactly opposite to) B.

   The magnetic Lorentz force on a charge q moving with velocity v in a field B is:

   F = q (v x B), with magnitude F = qvB sin(theta)

   where theta is the angle between v and B.

   For the proton to continue moving in a straight line along its initial direction (undeflected), the magnetic force on it must be zero at every instant, which requires sin(theta) = 0, i.e. theta = 0 degrees (v parallel to B) or theta = 180 degrees (v anti-parallel to B).

   …

   🔒 Start your 14-day free trial to see the full solution →
5. CBSE 2025Set ANNUAL1 mark
   Q.Write SI unit of magnetic flux density.

   ›Reveal solutionSolution

   Magnetic flux density (the magnetic field B) is measured in tesla.

   Magnetic flux density, commonly called the magnetic field B, is defined via the force on a moving charge, $F = qvB\sin\theta$, or via flux, $\Phi = BA\cos\theta$. Its SI unit is the tesla (T), where …

   🔒 Start your 14-day free trial to see the full solution →
6. CBSE 2025Set ANNUAL1 mark
   Q.A proton and alpha particle are moving northward with same velocity in a region of magnetic field of 3T directed downward. The magnetic force experienced by both proton and alpha particle. Which have greater magnetic force?

   ›Reveal solutionSolution

   $F=qvB\sin\theta$; same $v$, $B$, $\theta=90°$, but $q_\alpha=2q_p$.

   The magnetic force on a moving charge is $F=qvB\sin\theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$. Here both particles move north with the same velocity, and the field is directed downward, so $\theta=90°$ for both (velocity and field are perpendicular), and $F=qvB$ for each. Since the alpha particle has charge $+2e$ (twice the proton's charge $+e$), while $v$ and $B$ are identical for both: …

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2025Set ANNUAL1 mark
   Q.What is the work done by an external uniform magnetic field perpendicular to the velocity of a moving charge?

   ›Reveal solutionSolution

   The magnetic Lorentz force is perpendicular to velocity at every instant, so it can never do work.

   The force on a charge $q$ moving with velocity $\vec v$ in a magnetic field $\vec B$ is

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

   By the definition of the cross product, $\vec F$ is always perpendicular to $\vec v$. The work done in a small displacement $d\vec s = \vec v\,dt$ is

   $dW = \vec F \cdot d\vec s = q(\vec v\times\vec B)\cdot \vec v\,dt = 0$

   …

   🔒 Start your 14-day free trial to see the full solution →
8. CBSE 2025Set ANNUAL1 markMCQ
   Q.A charged particle is moving in a uniform magnetic field in a circular path of radius R. If the energy of the particle is doubled, then the new radius will be

   (a) R/√2

   (b) √2 R

   (c) 2R

   (d) 4R

   ›Reveal solutionSolution

   Radius of the circular path satisfies r ∝ √(kinetic energy), so doubling the energy scales r by √2.

   For a charged particle moving in a uniform magnetic field B, the magnetic force provides the centripetal force:

   $qvB=\dfrac{mv^2}{R}\ \Rightarrow\ R=\dfrac{mv}{qB}=\dfrac{p}{qB}$

   Since kinetic energy $E=\dfrac{p^2}{2m}$, we have $p=\sqrt{2mE}$, so …

   🔒 Start your 14-day free trial to see the full solution →
9. CBSE 2025Set ANNUAL1 mark
   Q.What is the magnitude of force experienced by a charge at rest placed in a uniform magnetic field? OR Between ammeter and voltmeter, which one has higher resistance?

   ›Reveal solutionSolution

   The magnetic force on a charge depends on its velocity; a charge at rest has v = 0, so the force is zero.

   The magnetic force on a moving charge is given by

   $\vec F = q\,\vec v \times \vec B$ …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2025Set ANNUAL1 mark
    Q.What is the force experienced by :

    (i) A stationary charge in a magnetic field ?

    (ii) A charge moving with a velocity v parallel to the magnetic field ?

    ›Reveal solutionSolution

    The magnetic force $F=qvB\sin\theta$ vanishes both when $v=0$ and when $v$ is parallel to $B$ ($\theta=0$).

    The magnetic force on a charge is $\vec F = q\vec v \times \vec B$, with magnitude $F = qvB\sin\theta$, $\theta$ being the angle between $\vec v$ and $\vec B$.

    (i) A stationary charge has $v=0$, so $F = qvB\sin\theta = 0$ regardless of the field - a magnetic field exerts no force on a charge at rest.

    …

    🔒 Start your 14-day free trial to see the full solution →
11. CBSE 2025Set ANNUAL1 markMCQ
    Q.When a charged particle moving with velocity v (vector) is subjected to a magnetic field of induction B (vector), the force on it is non zero. This implies that –

    (a) angle between them is either zero or 180°

    (b) angle between them is necessarily 90°

    (c) angle between them can have any value other than 90°

    (d) angle between them can have any value other than zero and 180°

    ›Reveal solutionSolution

    A non-zero magnetic force requires the velocity and field to not be parallel or antiparallel.

    The magnetic force on a charge is $\vec{F} = q\vec{v} \times \vec{B}$, with magnitude $F = qvB\sin\theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$.

    • If $\theta = 0°$ or $180°$, $\sin\theta = 0$, so $F = 0$. …

    🔒 Start your 14-day free trial to see the full solution →
12. CBSE 2024Set 55/5/11 markMCQ
    Q.Assertion (A): The energy of a charged particle moving in a magnetic field does not change. Reason (R): It is because the work done by the magnetic force on a charge moving in a magnetic field is zero. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is NOT the correct explanation of A. (C) A is true but R is false. (D) Both A and R are false.

    ›Reveal solutionSolution

    A magnetic force on a moving charge is always $\vec F=q\vec v\times\vec B$, which is perpendicular to $\vec v$ at every instant -- a force perpendicular to velocity does zero work, so the particle's kinetic energy (and hence its speed) never changes. Both A and R are true, and R is the correct explanation of A -- option (A).

    Why the magnetic force does no work

    The magnetic force on a charge $q$ moving with velocity $\vec v$ in a field $\vec B$ is

    $\vec F=q\,\vec v\times\vec B.$

    By the definition of the cross product, $\vec F$ is always perpendicular to $\vec v$, regardless of the instantaneous direction of $\vec v$ or $\vec B$.

    Power delivered by any force is $P=\vec F\cdot\vec v$. Since $\vec F\perp\vec v$ here, $\vec F\cdot\vec v=0$ at every instant, so

    $W=\int \vec F\cdot\vec v\,dt=0.$

    $\vec F = q\vec v\times\vec B \ \Rightarrow\ \vec F\perp\vec v \ \Rightarrow\ P=\vec F\cdot\vec v=0 \ \Rightarrow\ W=0.$ …

    🔒 Start your 14-day free trial to see the full solution →