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Fields - Motion in electromagnetic fields

Grade 12IBPhysics

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

🔑Concepts

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Magnetic Force on a Moving Charge: A particle with charge qq moving with velocity vv through a magnetic field BB experiences a force F=qvBsin⁡θF = qvB \sin\theta, where θ\theta is the angle between the velocity and the magnetic field vectors. The direction is determined by the Right-Hand Rule.

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Uniform Circular Motion: When a charged particle enters a uniform magnetic field perpendicularly (θ=90∘\theta = 90^\circ), the magnetic force acts as a centripetal force. This results in the particle following a circular path with a constant radius r=mvqBr = \frac{mv}{qB}.

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Motion in an Electric Field: A charged particle in a uniform electric field EE experiences a constant force F=qEF = qE. If the particle enters the field perpendicular to the field lines, it follows a parabolic trajectory, similar to a mass in a gravitational field.

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Crossed Fields (Velocity Selector): In a region with both electric and magnetic fields perpendicular to each other, a particle can move in a straight line if the electric force FE=qEF_E = qE and magnetic force FB=qvBF_B = qvB are equal and opposite. This occurs at a specific velocity v=EBv = \frac{E}{B}.

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Work and Energy: Magnetic forces do no work on a moving charge because the force is always perpendicular to the velocity (W=F⃗⋅d⃗=0W = \vec{F} \cdot \vec{d} = 0). However, electric fields do work on charges, changing their kinetic energy: W=qΔV=ΔEkW = q \Delta V = \Delta E_k.

📐Formulae

FB=qvBsin⁡θF_B = qvB \sin\theta

FE=qEF_E = qE

Fwire=BILsin⁡θF_{wire} = BIL \sin\theta

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

v=EBv = \frac{E}{B}

W=qV=12mv2−12mu2W = qV = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

💡Examples

Problem 1:

An electron (m=9.11×10−31 kgm = 9.11 \times 10^{-31} \text{ kg}, q=−1.60×10−19 Cq = -1.60 \times 10^{-19} \text{ C}) is accelerated from rest through a potential difference of 2000 V2000 \text{ V} and enters a uniform magnetic field of 0.015 T0.015 \text{ T} perpendicular to its motion. Calculate the radius of the electron's path.

Solution:

Step 1: Find the velocity vv of the electron using the conservation of energy. qV=12mv2qV = \frac{1}{2}mv^2 v=2qVmv = \sqrt{\frac{2qV}{m}} v=2×1.60×10−19×20009.11×10−31≈2.65×107 m s−1v = \sqrt{\frac{2 \times 1.60 \times 10^{-19} \times 2000}{9.11 \times 10^{-31}}} \approx 2.65 \times 10^7 \text{ m s}^{-1}

Step 2: Calculate the radius rr using the centripetal force relation. qvB=mv2rqvB = \frac{mv^2}{r} r=mvqBr = \frac{mv}{qB} r=(9.11×10−31)×(2.65×107)(1.60×10−19)×(0.015)r = \frac{(9.11 \times 10^{-31}) \times (2.65 \times 10^7)}{(1.60 \times 10^{-19}) \times (0.015)} r≈0.010 m=1.0 cmr \approx 0.010 \text{ m} = 1.0 \text{ cm}

Explanation:

The electric potential energy lost by the electron is converted into kinetic energy. Once in the magnetic field, the magnetic force acts as the centripetal force, causing the electron to move in a circle.

Problem 2:

A velocity selector is designed to allow protons to pass through undeflected when the electric field is 5.0×104 V m−15.0 \times 10^4 \text{ V m}^{-1}. Calculate the required magnetic field strength if the protons have a kinetic energy of 1.0×10−15 J1.0 \times 10^{-15} \text{ J}. (Mass of proton m=1.67×10−27 kgm = 1.67 \times 10^{-27} \text{ kg})

Solution:

Step 1: Calculate the velocity vv from the kinetic energy EkE_k. Ek=12mv2  ⟹  v=2EkmE_k = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2E_k}{m}} v=2×1.0×10−151.67×10−27≈1.1×106 m s−1v = \sqrt{\frac{2 \times 1.0 \times 10^{-15}}{1.67 \times 10^{-27}}} \approx 1.1 \times 10^6 \text{ m s}^{-1}

Step 2: Use the velocity selector condition where the forces balance. qE=qvB  ⟹  B=EvqE = qvB \implies B = \frac{E}{v} B=5.0×1041.1×106B = \frac{5.0 \times 10^4}{1.1 \times 10^6} B≈0.045 TB \approx 0.045 \text{ T}

Explanation:

In a velocity selector, only particles with velocity v=EBv = \frac{E}{B} experience zero net force because the upward electric force cancels the downward magnetic force (or vice versa).