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#charged particle

9 public questions tagged with this topic.

Equipotential Surfaces and Relation Between Field and Potential

A charged particle is placed in a region where the electric potential is constant everywhere. What can be inferred about

**Relation E = -∇V** shows field points down potential gradient. For system of opposite charges close together, equipotential near midpoint between them has V≈0, but shape distorted, not spherical, reflecting superposition of potentials V = k q₁/r₁ + k q₂/r₂. If the electric potential V is constant everywhere, the electric field E = -(dV/dx) = 0 , since the gradient of a constant potential is zero. With no electric field, there is no electrostatic force ( F = qE ) acting on the charged particle. If released from rest, the particle experiences no acceleration due to the electric field and thus

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Equipotential Surfaces and Relation Between Field and Potential

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Solenoid, Toroid and Ampere's Law

An electron moves at \( 5 \times 10^6 \, \text{m/s} \) perpendicular to a magnetic field of \( 0.2 \, \text{T} \). What

**Solenoid field** inside long solenoid is B = μ₀ n I, n = N/L turns per meter (m⁻¹), uniform and parallel to axis, outside negligible for long solenoid because fields from opposite sides cancel. For n = 1200 m⁻¹, I = 1 A, B = 4π×10⁻⁷×1200 = 1.51×10⁻³ T = 1.51 mT. Force F = q v B sin θ , θ = 90° , so sin θ = 1 . F = 1.6 × 10⁻¹⁹ × 5 × 10⁶ × 0.2 = 1.6 × 10⁻¹³ N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Solenoid, Toroid and Ampere's Law

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A proton moves at \( 2.5 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.15 \, \text{T} \). What is the ma

**Motion of charged particle perpendicular to uniform B** is circular because force F ⊥ v, no work done, speed constant, radius r = m v/(q B), period T = 2π m/(q B) independent of v. If v has component parallel to B, helical path results, pitch = v_parallel·T. Force F = q v B sin θ , θ = 90° , so sin θ = 1 . F = 1.6 × 10⁻¹⁹ × 2.5 × 10⁷ × 0.15 = 6 × 10⁻¹³ N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A proton moves with a speed of \( 1.8 \times 10^6 \, \text{m/s} \) perpendicular to a magnetic field of \( 0.5 \, \text{

**Effect of doubling velocity** on magnetic force F = q v B sinθ is linear increase, F doubles for same θ and B. Electron with charge 1.6×10⁻¹⁹ C, v = 4.5×10⁶ m/s, B = 0.35 T, θ = 90°, F = 1.6×10⁻¹⁹×4.5×10⁶×0.35 = 2.52×10⁻¹³ N, illustrating magnitude for typical lab values. Radius r = (mv/qB) . r = (1.67 × 10⁻²⁷ × 1.8 × 10⁶/1.6 × 10⁻¹⁹ × 0.5) = (3.006 × 10⁻²¹/8 × 10⁻²⁰) = 3.7575 × 10⁻² m = 3.76 cm . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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Magnetic Force on Moving Charge - Lorentz Force and Motion

An electron moves with a speed of \( 3.5 \times 10^6 \, \text{m/s} \) perpendicular to a magnetic field of \( 0.25 \, \t

**Lorentz force** on charge q moving with velocity v in magnetic field B is F = q v × B, magnitude F = q v B sinθ, θ angle between v and B (degrees), unit N. Direction perpendicular to both v and B via right-hand rule. When v ⊥ B, motion circular with radius r = m v/(q B), centripetal force provided by magnetic force. Radius r = (mv/qB) . r = (9.1 × 10⁻³¹ × 3.5 × 10⁶/1.6 × 10⁻¹⁹ × 0.25) = (3.185 × 10⁻²⁴/4 × 10⁻²⁰) = 7.9625 × 10⁻⁵ m = 7.96 × 10⁻³ cm . Using F = q v

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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Magnetic Force on Moving Charge - Lorentz Force and Motion

Why does a charged particle describe a circular path when moving perpendicular to a uniform magnetic field?

**Effect of doubling velocity** on magnetic force F = q v B sinθ is linear increase, F doubles for same θ and B. Electron with charge 1.6×10⁻¹⁹ C, v = 4.5×10⁶ m/s, B = 0.35 T, θ = 90°, F = 1.6×10⁻¹⁹×4.5×10⁶×0.35 = 2.52×10⁻¹³ N, illustrating magnitude for typical lab values. The magnetic force F = q (v × B) acts perpendicular to the velocity, providing the centripetal force needed for circular motion when the velocity is perpendicular to the field. Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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Magnetic Force on Moving Charge - Lorentz Force and Motion

What is the effect on the magnetic force if a charged particle’s velocity is doubled while keeping the magnetic field co

**Effect of doubling velocity** on magnetic force F = q v B sinθ is linear increase, F doubles for same θ and B. Electron with charge 1.6×10⁻¹⁹ C, v = 4.5×10⁶ m/s, B = 0.35 T, θ = 90°, F = 1.6×10⁻¹⁹×4.5×10⁶×0.35 = 2.52×10⁻¹³ N, illustrating magnitude for typical lab values. The magnetic force F = q v B sin θ is directly proportional to velocity v . Doubling the velocity doubles the force. Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N I A B sinθ,

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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Torque on Magnetic Dipole and Potential Energy

The reason a magnetic field does not exert a force along its own direction on a moving charge is:

**Potential energy of magnetic dipole** U = -m B cosθ explains stability. Given m = 0.9 A·m², B = 0.5 T, θ = 90°, sin90° = 1, τ = 0.45 N·m. For 60°, sin60° = √3/2 ≈0.866, reducing torque proportionally. The magnetic force on a moving charge is given by F = q (v × B) , which is always perpendicular to both the velocity v and the magnetic field B . Thus, the force has no component along the field direction, as the cross product ensures orthogonality. Substituting values gives The force is perpendicular to the field, which matches expected magnitude for this magnetic

Ref: NCERT > Physics Book > Magnetism and Matter > Torque on Magnetic Dipole and Potential Energy

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