krit.club logo

Physics - Electricity and Magnetism (Static Electricity, Current, Simple Circuits, Magnetic Field)

Grade 9ICSE

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

🔑Concepts

•

Static Electricity: This involves the study of electric charges at rest. Charging occurs due to the transfer of electrons (e−e^-). A body becomes positively charged by losing electrons and negatively charged by gaining them.

•

Coulomb's Law: The force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

•

Electric Current (II): The rate of flow of electric charge through a conductor, measured in Amperes (AA). 1 A=1 C s−11 \text{ A} = 1 \text{ C s}^{-1}.

•

Potential Difference (VV): The amount of work done in moving a unit positive charge from one point to another in an electric field. Measured in Volts (VV).

•

Resistance (RR): The opposition offered by a conductor to the flow of current. According to Ohm's Law, at constant temperature, V∝IV \propto I.

•

Magnetic Field: The space around a magnet where its influence can be detected. Magnetic field lines emerge from the North pole and enter the South pole outside the magnet.

•

Electromagnetism: A current-carrying conductor produces a magnetic field around it. The direction is determined by the Right-Hand Thumb Rule.

•

Induced Magnetism: A temporary magnetism acquired by a magnetic material (like soft iron) when placed near a magnet.

📐Formulae

Q=neQ = ne

I=QtI = \frac{Q}{t}

V=WQV = \frac{W}{Q}

V=IRV = IR

R=ρlAR = \rho \frac{l}{A}

💡Examples

Problem 1:

Calculate the current flowing through a wire if a charge of 120 C120 \text{ C} flows through it in 22 minutes.

Solution:

Given: Q=120 CQ = 120 \text{ C}, t=2 minutes=2×60=120 st = 2 \text{ minutes} = 2 \times 60 = 120 \text{ s}. Using the formula I=QtI = \frac{Q}{t}, we get I=120120=1 AI = \frac{120}{120} = 1 \text{ A}.

Explanation:

To find the current, the time must be converted into the SI unit (seconds) before dividing the total charge by the time duration.

Problem 2:

How much work is done in moving a charge of 5 C5 \text{ C} across two points having a potential difference of 12 V12 \text{ V}?

Solution:

Given: Q=5 CQ = 5 \text{ C}, V=12 VV = 12 \text{ V}. Using the formula V=WQV = \frac{W}{Q}, we rearrange to find W=V×QW = V \times Q. Thus, W=12×5=60 JW = 12 \times 5 = 60 \text{ J}.

Explanation:

Work done is the product of the potential difference and the magnitude of the charge moved.

Problem 3:

A conductor has a resistance of 10 Ω10 \, \Omega. If a potential difference of 5 V5 \text{ V} is applied across its ends, calculate the current.

Solution:

Given: R=10 ΩR = 10 \, \Omega, V=5 VV = 5 \text{ V}. From Ohm's Law, I=VRI = \frac{V}{R}. So, I=510=0.5 AI = \frac{5}{10} = 0.5 \text{ A}.

Explanation:

Current is directly proportional to potential difference and inversely proportional to resistance as per V=IRV = IR.