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Fields - Magnetic Fields

Grade 11IBPhysics

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

🔑Concepts

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Magnetic fields (denoted by BB) are vector fields that exert forces on moving charges and current-carrying conductors. The SI unit for magnetic field strength is the Tesla (TT).

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Magnetic field lines represent the direction and magnitude of the field. They always point from the North pole to the South pole outside a magnet and their density indicates the field strength.

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A point charge qq moving with velocity vv in 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.

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A straight conductor of length LL carrying a current II in a magnetic field experiences a force F=BILsin⁡θF = BIL \sin\theta.

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The direction of the magnetic force is determined by the Right-Hand Rule: for a positive charge, if the thumb points in the direction of velocity vv and the fingers in the direction of BB, the palm points in the direction of the force FF.

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Since the magnetic force is always perpendicular to the velocity of a charged particle, it does no work and results in uniform circular motion with a radius r=mvqBr = \frac{mv}{qB}.

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The magnetic field strength BB at a distance rr from a long straight current-carrying wire is given by B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}, where μ0\mu_0 is the permeability of free space (4π×10−7 T m A−14\pi \times 10^{-7} \text{ T m A}^{-1}).

📐Formulae

F=qvBsin⁡θF = qvB \sin\theta

F=BILsin⁡θF = BIL \sin\theta

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

B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}

Φ=BAcos⁡θ\Phi = BA \cos\theta

💡Examples

Problem 1:

An electron (q=−1.6×10−19 Cq = -1.6 \times 10^{-19} \text{ C}) enters a uniform magnetic field of 0.25 T0.25 \text{ T} at a velocity of 4.0×106 m s−14.0 \times 10^6 \text{ m s}^{-1} perpendicular to the field lines. Calculate the magnitude of the magnetic force acting on the electron.

Solution:

Using the formula F=qvBsin⁡θF = qvB \sin\theta: F=(1.6×10−19 C)×(4.0×106 m s−1)×(0.25 T)×sin⁡(90∘)F = (1.6 \times 10^{-19} \text{ C}) \times (4.0 \times 10^6 \text{ m s}^{-1}) \times (0.25 \text{ T}) \times \sin(90^\circ) F=1.6×10−13 NF = 1.6 \times 10^{-13} \text{ N}.

Explanation:

Since the velocity is perpendicular to the field, θ=90∘\theta = 90^\circ and sin⁡(90∘)=1\sin(90^\circ) = 1. The magnitude of the charge is used for the force magnitude calculation.

Problem 2:

A wire of length 0.10 m0.10 \text{ m} carries a current of 5.0 A5.0 \text{ A} in a direction that makes an angle of 30∘30^\circ with a uniform magnetic field of 0.40 T0.40 \text{ T}. Determine the force on the wire.

Solution:

F=BILsin⁡θF = BIL \sin\theta F=(0.40 T)×(5.0 A)×(0.10 m)×sin⁡(30∘)F = (0.40 \text{ T}) \times (5.0 \text{ A}) \times (0.10 \text{ m}) \times \sin(30^\circ) F=0.40×5.0×0.10×0.5F = 0.40 \times 5.0 \times 0.10 \times 0.5 F=0.10 NF = 0.10 \text{ N}.

Explanation:

The force is calculated using the component of the magnetic field perpendicular to the current, which is represented by sin⁡(30∘)=0.5\sin(30^\circ) = 0.5.

Magnetic Fields Grade 11 Notes & Examples | IB Physics