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Moving Charges and Magnetism - Force on a Moving Charge in Magnetic Field

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Magnetic Lorentz Force: When a charge qq moves with velocity v⃗\vec{v} in a magnetic field B⃗\vec{B}, it experiences a force perpendicular to both v⃗\vec{v} and B⃗\vec{B}.

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The magnitude of the force is maximum when the velocity is perpendicular to the magnetic field (θ=90∘\theta = 90^\circ) and zero when moving parallel or anti-parallel to the field (θ=0∘\theta = 0^\circ or 180∘180^\circ).

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Direction of force: For a positive charge, the direction is given by the Right-Hand Rule (cross product v⃗×B⃗\vec{v} \times \vec{B}) or Fleming's Left-Hand Rule. For a negative charge, the direction is reversed.

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Work Done and Energy: Since the force F⃗m\vec{F}_m is always perpendicular to velocity v⃗\vec{v}, the work done by the magnetic field on the charge is zero (W=∫F⃗⋅ds⃗=0W = \int \vec{F} \cdot d\vec{s} = 0). Consequently, the kinetic energy and speed of the particle remain constant, though its direction changes.

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Circular Motion: If v⃗⊥B⃗\vec{v} \perp \vec{B}, the particle undergoes uniform circular motion where the magnetic force provides the necessary centripetal force.

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Helical Motion: If the velocity v⃗\vec{v} makes an angle θ\theta (where 0<θ<90∘0 < \theta < 90^\circ) with B⃗\vec{B}, the velocity component vsin⁡θv \sin \theta causes circular motion while vcos⁡θv \cos \theta causes linear motion, resulting in a helical path.

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Lorentz Force: The total force on a charge moving in a region with both electric field E⃗\vec{E} and magnetic field B⃗\vec{B} is F⃗=q(E⃗+v⃗×B⃗)\vec{F} = q(\vec{E} + \vec{v} \times \vec{B}).

📐Formulae

F⃗m=q(v⃗×B⃗)\vec{F}_m = q(\vec{v} \times \vec{B})

F=qvBsin⁡θF = qvB \sin \theta

r=mvqB=pqB=2mKqBr = \frac{mv}{qB} = \frac{p}{qB} = \frac{\sqrt{2mK}}{qB}

T=2πmqBT = \frac{2\pi m}{qB}

f=qB2πmf = \frac{qB}{2\pi m}

Pitch=(vcos⁡θ)T=2πmvcos⁡θqB\text{Pitch} = (v \cos \theta) T = \frac{2\pi m v \cos \theta}{qB}

F⃗net=qE⃗+q(v⃗×B⃗)\vec{F}_{net} = q\vec{E} + q(\vec{v} \times \vec{B})

💡Examples

Problem 1:

An alpha particle (q=+2eq = +2e, m=6.64×10−27 kgm = 6.64 \times 10^{-27}\text{ kg}) enters a uniform magnetic field of 2.0 T2.0\text{ T} with a velocity of 3×106 m/s3 \times 10^6\text{ m/s} perpendicular to the field. Calculate the radius of its path.

Solution:

Given: q=2×1.6×10−19 Cq = 2 \times 1.6 \times 10^{-19}\text{ C}, B=2.0 TB = 2.0\text{ T}, v=3×106 m/sv = 3 \times 10^6\text{ m/s}, m=6.64×10−27 kgm = 6.64 \times 10^{-27}\text{ kg}. Using r=mvqBr = \frac{mv}{qB}: r=(6.64×10−27)×(3×106)(3.2×10−19)×2.0r = \frac{(6.64 \times 10^{-27}) \times (3 \times 10^6)}{(3.2 \times 10^{-19}) \times 2.0} r=1.992×10−206.4×10−19≈0.0311 m=3.11 cmr = \frac{1.992 \times 10^{-20}}{6.4 \times 10^{-19}} \approx 0.0311\text{ m} = 3.11\text{ cm}

Explanation:

Since the velocity is perpendicular to the magnetic field, the particle follows a circular path. The magnetic force qvBqvB provides the centripetal force mv2r\frac{mv^2}{r}.

Problem 2:

A proton is moving along the positive xx-axis in a region where a magnetic field points along the positive zz-axis. What is the direction of the magnetic force?

Solution:

Velocity v⃗=vi^\vec{v} = v\hat{i}, Magnetic field B⃗=Bk^\vec{B} = B\hat{k}. Force F⃗=q(v⃗×B⃗)\vec{F} = q(\vec{v} \times \vec{B}). For a proton, q>0q > 0. F⃗=q(vi^×Bk^)=qvB(i^×k^)=qvB(−j^)\vec{F} = q(v\hat{i} \times B\hat{k}) = qvB(\hat{i} \times \hat{k}) = qvB(-\hat{j}) The force is directed along the negative yy-axis.

Explanation:

Using the cross product rule for unit vectors, i^×k^=−j^\hat{i} \times \hat{k} = -\hat{j}. Alternatively, using Fleming's Left-Hand Rule: Forefinger (Field) points up (+z+z), Middle finger (Current/Velocity) points right (+x+x), Thumb (Force) points down (−y-y).