krit.club logo

Fields - Electric and magnetic fields

Grade 12IBPhysics

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

🔑Concepts

•

An electric field is a region of space where a charged object experiences a force. The electric field strength EE is defined as the force per unit charge acting on a small positive test charge: E=FqE = \frac{F}{q}.

•

Coulomb's Law states that the electrostatic force FF between two point charges is proportional to the product of the charges q1q_1 and q2q_2 and inversely proportional to the square of the distance rr between them: F=kq1q2r2F = k \frac{q_1 q_2}{r^2}.

•

Electric field lines represent the path a positive test charge would follow. They point away from positive charges and toward negative charges.

•

Magnetic fields are created by moving charges (currents) or permanent magnets. The direction of the magnetic field BB is the direction that the North pole of a small compass needle points.

•

The magnetic force FF on a charge qq moving with velocity vv at an angle θ\theta to a magnetic field BB is given by F=qvBsin⁡θF = qvB\sin\theta. The force is always perpendicular to both the velocity and the magnetic field (Right-Hand Rule).

•

A current-carrying conductor in a magnetic field experiences a force F=BILsin⁡θF = BIL\sin\theta, where II is the current, LL is the length of the wire, and BB is the magnetic field strength.

📐Formulae

E=FqE = \frac{F}{q}

F=kq1q2r2F = k \frac{q_1 q_2}{r^2}

E=kQr2E = k \frac{Q}{r^2}

k=14πϵ0k = \frac{1}{4\pi\epsilon_0}

F=qvBsin⁡θF = qvB\sin\theta

F=BILsin⁡θF = BIL\sin\theta

💡Examples

Problem 1:

Determine the magnitude of the electric field strength at a point 2.0 m2.0 \text{ m} away from a point charge of +5.0μC+5.0 \mu\text{C}. Take k=8.99×109 N m2 C−2k = 8.99 \times 10^9 \text{ N m}^2 \text{ C}^{-2}.

Solution:

Using the formula for the electric field of a point charge: E=kQr2E = k \frac{Q}{r^2} E=(8.99×109)×5.0×10−6(2.0)2E = (8.99 \times 10^9) \times \frac{5.0 \times 10^{-6}}{(2.0)^2} E=(8.99×109)×5.0×10−64.0E = (8.99 \times 10^9) \times \frac{5.0 \times 10^{-6}}{4.0} E=1.12375×104 N C−1E = 1.12375 \times 10^4 \text{ N C}^{-1}

Explanation:

The electric field strength follows an inverse square law relative to the distance from the point charge.

Problem 2:

An electron (q=−1.6×10−19 Cq = -1.6 \times 10^{-19} \text{ C}) moves at a speed of 2.5×106 m s−12.5 \times 10^6 \text{ m s}^{-1} perpendicular to a uniform magnetic field of 0.40 T0.40 \text{ T}. Calculate the magnetic force acting on the electron.

Solution:

Since the motion is perpendicular, θ=90∘\theta = 90^{\circ} and sin⁡(90∘)=1\sin(90^{\circ}) = 1. F=qvBF = qvB F=(1.6×10−19)×(2.5×106)×0.40F = (1.6 \times 10^{-19}) \times (2.5 \times 10^6) \times 0.40 F=1.6×10−13 NF = 1.6 \times 10^{-13} \text{ N}

Explanation:

The force is calculated using the magnitude of the charge. The direction of the force would be determined by the Right-Hand Rule, reversing the result for a negative charge.