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Magnetic Effects of Electric Current - 12.212.2 12.212.2 MAMAMA

Grade 10CBSE

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

🔑Concepts

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Magnetic Field and Field Lines: A magnetic field is a region around a magnet where magnetic force can be experienced. Field lines emerge from the North pole and merge at the South pole. Inside the magnet, the direction is from South to North, forming closed curves.

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Right-Hand Thumb Rule: Used to find the direction of the magnetic field around a straight current-carrying conductor. If the thumb points in the direction of current II, the curled fingers show the direction of the magnetic field BB.

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Magnetic Field due to a Solenoid: A solenoid is a coil of many circular turns of insulated copper wire. The magnetic field inside a long solenoid is uniform and represented by parallel straight lines. B∝nIB \propto nI, where nn is the number of turns per unit length.

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Fleming's Left-Hand Rule: Used to find the direction of force FF on a current-carrying conductor in a magnetic field BB. Stretch the thumb, forefinger, and middle finger of the left hand mutually perpendicular. Forefinger = Field, Middle finger = Current, Thumb = Motion/Force.

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Electromagnetic Induction (EMI): The process by which a changing magnetic field in a conductor induces a current in another conductor. This is the principle behind electric generators.

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Fleming's Right-Hand Rule: Used to find the direction of induced current. Thumb = Direction of motion of conductor, Forefinger = Magnetic field, Middle finger = Induced current.

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Factors affecting the magnetic field of a circular loop: BB is directly proportional to the current II and the number of turns NN, and inversely proportional to the radius rr (B∝NIrB \propto \frac{NI}{r}).

📐Formulae

B∝IB \propto I

B∝1rB \propto \frac{1}{r}

F=BIlsin⁡θF = BIl \sin \theta

B=μ0nI (for a solenoid)B = \mu_0 n I \text{ (for a solenoid)}

Φ=B⋅Acos⁡θ (Magnetic Flux)\Phi = B \cdot A \cos \theta \text{ (Magnetic Flux)}

💡Examples

Problem 1:

A current-carrying conductor of length l=0.5 ml = 0.5\text{ m} is placed perpendicular to a magnetic field of B=2 TB = 2\text{ T}. If the current flowing through it is I=1.5 AI = 1.5\text{ A}, calculate the force acting on the conductor.

Solution:

Given: l=0.5 ml = 0.5\text{ m}, B=2 TB = 2\text{ T}, I=1.5 AI = 1.5\text{ A}, and θ=90∘\theta = 90^\circ (perpendicular). Using the formula: F=BIlsin⁡90∘F = BIl \sin 90^\circ Since sin⁡90∘=1\sin 90^\circ = 1: F=2×1.5×0.5F = 2 \times 1.5 \times 0.5 F=1.5 NF = 1.5\text{ N}

Explanation:

The force on a conductor is maximum when it is placed perpendicular to the magnetic field. We substitute the values into the Lorentz force equation for a wire.

Problem 2:

An electron enters a magnetic field at right angles to it as shown in a diagram. The direction of the magnetic field is into the page and the electron moves from left to right. Determine the direction of the force.

Solution:

Direction of current II is opposite to the flow of electrons (Right to Left). Field BB is into the page. Using Fleming's Left-Hand Rule: Forefinger (Field) points into the page, Middle finger (Current) points Left. The Thumb points downwards.

Explanation:

According to Fleming's Left-Hand Rule, the force acting on the electron will be towards the bottom of the page.