Grade 10 Physics Review Notes

Magnetic Fields and Forces

• A current-carrying straight wire passes through a 10T uniform magnetic field. The length of the wire in the magnetic field is 3m, and the current direction makes a 30-degree angle with the field. If the wire experiences a 45N force, the current through the wire can be calculated using the formula: $F = BIL {sin}(\theta)$, where $F$ is the force, $B$ is the magnetic field strength, $I$ is the current, $L$ is the length of the wire in the field, and ${θ}$ is the angle between the current and the field. Therefore, $I = F / (BL {sin}(\theta)) = 45 / (10 * 3 * {sin}(30)) = 3A$.

• A particle with a mass of $3.2 × 10^{-9}$ kg and a velocity of $5 × 10^3$ m/s has a charge of 12μC and moves in a radius of 100m. The magnetic field strength $B$ can be found using the formula: $B = (mv) / (qR)$, where $m$ is the mass, $v$ is the velocity, $q$ is the charge, and $R$ is the radius. Thus, $B = (3.2 × 10^{-9} * 5 × 10^3) / (12 × 10^{-6} * 100) ≈ 0.00133 {T}$.

• A particle of mass $m$ fired into a magnetic field of strength $B$ at speed $v$ travels in a circular path of radius $r$. The magnitude of the charge on the particle can be expressed as: $q = (mv) / (Br)$.

Electrical Circuits and Resistance

• Ohm's Law: $V = IR$, where $V$ is voltage, $I$ is current, and $R$ is resistance.

• Resistors in Series: The total resistance ($R {total}$) is the sum of individual resistances: $R {total} = R<em>1 + R</em>2 + R_3 + …$

• Resistors in Parallel: The reciprocal of the total resistance is the sum of the reciprocals of individual resistances: ${1/R {total}} = 1/R<em>1 + 1/R</em>2 + 1/R_3 + …$

• A boy with skin resistance of 100,000 Ω touches a 7V battery. The current flow through the boy is: $I = V/R = 7 / 100000 = 7 × 10^{-5} A$.

• Constantan wire with resistivity of $4.9 × 10^{-7} Ωm$, length of 3m, and cross-sectional area of $1 {mm}^2$ ($1 × 10^{-6} m^2$) has a resistance of: $R = ρ(L/A) = (4.9 × 10^{-7} * 3) / (1 × 10^{-6}) = 1.47 Ω$.

• To calculate the effective resistance of a circuit with three resistors connected between points a and b, the configuration (series or parallel) must be known.

• The charge transported by 15A of current in one hour is: $Q = I * t = 15 * (1 * 3600) = 54000 C$.

• A 34V battery connected to a resistive lamp with a current of 12A has a lamp resistance of: $R = V/I = 34 / 12 ≈ 2.83 Ω$.

• $1.35 × 10^{24}$ electrons passing through a conductor in 2 hours (7200 s) results in a current of: $I = (n * e) / t = (1.35 × 10^{24} * 1.6 × 10^{-19}) / 7200 ≈ 30 A$.

• A silver wire with a length of 180m and a cross-sectional area of $0.3 {mm}^2$ ($0.3 × 10^{-6} m^2$) and resistivity of $1.6 × 10^{-8} Ωm$ has a resistance of: $R = ρ(L/A) = (1.6 × 10^{-8} * 180) / (0.3 × 10^{-6}) = 9.6 Ω$.

• A resistor $R$ connected with a 15Ω resistance to provide an effective resistance of 6Ω implies a parallel connection. $\frac{1}{6} = \frac{1}{15} + \frac{1}{R}$. Solving for R gives $R = 10\Omega$. The resistors are connected in parallel.

• A voltage needed to drive a current of 0.8A through a torch lump of resistance 12 Ω is: $V = I * R = 0.8 * 12 = 9.6 V$.

• A copper wire with a resistance of 4 Ω, when its cross-section is doubled and length halved, the new resistance is $R {new} = ρ((L/2) / (2A)) = (1/4) * ρ(L/A) = (1/4) * 4 = 1 Ω$.

Electric Fields and Forces

• A charge of 8 μC experiencing a force of $4 × 10^{-4}$ N has an electric field strength of: $E = F/q = (4 × 10^{-4}) / (8 × 10^{-6}) = 50 N/C$.

• Two charges, $Q<em>1 = 2 × 10^{-6} C$ and $Q</em>2 = 5 × 10^{-6} C$, exerting a force of 9N on each other are separated by a distance of: $F = K * (Q<em>1 * Q</em>2) / r^2$, where $K = 9 × 10^9 Nm^2/C^2$. Solving for $r$ gives $r = \sqrt{(K * Q<em>1 * Q</em>2) / F} = \sqrt{((9 × 10^9) * (2 × 10^{-6}) * (5 × 10^{-6})) / 9} = 0.1 m$.

• An iron railway line with resistivity of $1 × 10^{-7} Ωm$, length of 12000m, and cross-sectional area of $200 {cm}^2$ ($0.02 m^2$) has a resistance of: $R = ρ(L/A) = (1 × 10^{-7} * 12000) / 0.02 = 0.06 Ω$.

• The net force acting on the -2 μC charge due to the other two charges can be calculated using Coulomb's Law for each pair and then summing the forces vectorially.

Magnetism

• A magnet is an object that produces a magnetic field.

  • Magnetic field lines never intersect.

  • They form closed loops.

  • They point from the north pole to the south pole outside the magnet.

• A magnet loses its magnetic properties when heated above its Curie temperature or subjected to a strong opposing magnetic field.

• Like magnetic poles repel, and unlike magnetic poles attract.

• A temporary magnet is easily magnetized and demagnetized, while a permanent magnet retains its magnetism.

• The magnetic flux when a magnetic field strength of 30T covers an area of $400 {cm}^2$ ($0.04 m^2$) is: $Φ = B * A = 30 * 0.04 = 1.2 Wb$.

• The direction of the magnetic force between two parallel current-carrying conductors depends on the direction of the currents.

  • If currents are in the same direction, the force is attractive.

  • If currents are in opposite directions, the force is repulsive.

• An electron moving at $10 × 10^4$ m/s in a 22T magnetic field experiences a force of: $F = qvB = (1.6 × 10^{-19}) * (10 × 10^4) * 22 = 3.52 × 10^{-13} N$.

• For a particle moving in a magnetic field, the magnetic field strength is: $B = (mv) / (qR) = (2 × 10^{-27} * 6 × 10^6) / (3 * 100) = 4 × 10^{-29} T$.

• A straight wire of length 50cm (0.5m) carrying a current of 1.2A in a 2T magnetic field experiences a force of: $F = BIL = 2 * 1.2 * 0.5 = 1.2 N$.

• A wire 25 cm long is at right angles to a 0.30T uniform magnetic field. The current through the wire is 6.0 A. The magnitude of the force on the wire: $F = B<em>I</em>L = 0.3<em>6</em>0.25 = 0.45 N$.

• A wire 0.50m long carrying a current of 8.0 A is at right angles to a uniform magnetic field. The force on the wire is 0.40 N. The strength of the magnetic field: $B = F/(I<em>L) = 0.40/(8</em>0.5) = 0.1 T$.

Factors Affecting Magnetic Force on Current-Carrying Wire

• Magnetic field strength (B).

• Current (I).

• Length of the wire (L).

• Angle between the wire and the magnetic field (θ).

Right-Hand Rule

• Point your thumb in the direction of the current.

• Your fingers curl in the direction of the magnetic field.

Additional Magnetism Challenges

• A wire of length 400m in a 0.20 T magnetic field experiences a 2.5 N force, the current is: $I = F / (BL) = 2.5 / (0.20 * 400) = 0.03125 A$.

Optics and Wave Phenomena

• When an object 30cm in front of a concave mirror with a focal length of 50cm, the image location can be found using the mirror equation: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$. Where $f = 50 cm$, $u = -30 cm$. $\frac{1}{50} = \frac{1}{v} + \frac{1}{-30}$. Therefore, $v = -75 cm$. The image is virtual, located 75 cm behind the mirror.

• The image formed by a concave mirror with a radius of curvature of 40cm is formed at half the radius of curvature, which is the focal point. Therefore, the image is formed 20cm from the mirror.

• If the radius of curvature of the concave mirror is 200m, and the size of the object twice as large as the object, then the image is formed at $f = R/2$, the image is formed 100m from the mirror.

• The main cause of a rainbow is dispersion and refraction of sunlight by water droplets in the atmosphere during rain.

• Light incident from air into glass medium of index of refraction $n = 8/5$. If the speed of light in air is $3 × 10^8$ m/s, its speed in the glass medium is: $v = c/n = (3 × 10^8) / (8/5) = 1.875 × 10^8 m/s$.

• The critical angle is the angle of incidence beyond which total internal reflection occurs.

• Total internal reflection is the phenomenon where light is entirely reflected at the boundary when it travels from a denser medium to a rarer medium at an angle of incidence greater than the critical angle.

• The speed of light in one type of glass is $2 × 10^8 m/s$. Its refractive index is: $n = c/v = (3 × 10^8) / (2 × 10^8) = 1.5$.

• Reflection is the bouncing back of light from a surface, while refraction is the bending of light as it passes from one medium to another.

• Radio program is broadcast in medium wave band with wave length 500m, the velocity of radio wave is $3x10^8m/s$, then the frequency corresponding to this wave is $f = v/\lambda = (3<em>10^8)/500 = 6</em>10^5 Hz$

Plane Mirror Image Characteristics

• The image is laterally inverted.

• The image is virtual.

• The image is the same size as the object.

• The image is upright.

Snell's Law

• Snell's Law states the relationship between the angles of incidence and refraction when light passes between two different media:
  $n<em>1 \sin(\theta</em>1) = n<em>2 \sin(\theta</em>2)$

Concave Mirror Image Formation

• When an object is placed at the center of curvature of a concave mirror, the image is formed at the center of curvature.

• The image is real, inverted, and the same size as the object.

Vision Defects and Correction

• Myopia (nearsightedness): Corrected with concave lenses.

• Hyperopia (farsightedness): Corrected with convex lenses.

• Astigmatism: Corrected with cylindrical lenses.

Primary Colors

• The addition of primary colors (red, green, blue) produces white light.

Dispersion

• Dispersion is the splitting of white light into its constituent colors.

Reflection Types

• Specular reflection: Reflection from a smooth surface.

• Diffuse reflection: Reflection from a rough surface.

Mirror and Refraction Calculations

• A mirror has a focal length of 0.3m. If an object is placed 0.6m from the mirror, the image is formed at: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$. Where $f = 0.3 m$, $u = -0.6 cm$. $\frac{1}{0.3} = \frac{1}{v} + \frac{1}{-0.6}$. Therefore, $v = 0.6 m$.

• A ray of light in air strikes the surface of a liquid at an angle of $60^0$ with the normal. The refracted ray is at an angle of $30^0$ with the normal. The index of refraction of this liquid is: $n = \frac{\sin(\theta<em>1)}{\sin(\theta</em>2)} = \frac{\sin(60)}{\sin(30)} = \sqrt{3} ≈ 1.732$.

• An object 2cm high is placed 5cm in front of a concave mirror with a focal length of 10cm. the image located at: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$. Where $f = 10 cm$, $u = -5 cm$. $\frac{1}{10} = \frac{1}{v} + \frac{1}{-5}$. Therefore, $v = -10 cm$. The image is virtual, located 10 cm behind the mirror.

• A concave mirror has a radius of curvature of 30cm and is positioned such that the upright image of an object is 2 times the size of the object. How far is the object from the mirror? $R = 30 cm$, so $f = R/2 = 15 cm$. Magnification $M = -v/u = 2$, so $v = -2u$. Using the mirror equation: $\frac{1}{15} = \frac{1}{-2u} + \frac{1}{u}$. Therefore, $u = 7.5 cm$.