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Fields - Magnetic Fields

Grade 11IBPhysics

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

🔑Concepts

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).

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.

A point charge qq moving with velocity vv in a magnetic field BB experiences a force F=qvBsinθF = qvB \sin\theta, where θ\theta is the angle between the velocity and the magnetic field vectors.

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

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.

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}.

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×107 T m A14\pi \times 10^{-7} \text{ T m A}^{-1}).

📐Formulae

F=qvBsinθF = qvB \sin\theta

F=BILsinθF = BIL \sin\theta

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

B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}

Φ=BAcosθ\Phi = BA \cos\theta

💡Examples

Problem 1:

An electron (q=1.6×1019 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 s14.0 \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=qvBsinθF = qvB \sin\theta: F=(1.6×1019 C)×(4.0×106 m s1)×(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=1.6×1013 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 3030^\circ with a uniform magnetic field of 0.40 T0.40 \text{ T}. Determine the force on the wire.

Solution:

F=BILsinθF = 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 - Revision Notes & Key Formulas | IB Grade 11 Physics