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Electromagnetic Induction - Magnetic Flux

Grade 12CBSEPhysics

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

🔑Concepts

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Magnetic flux ΦB\Phi_B linked with a surface is defined as the total number of magnetic field lines passing normally through that surface.

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It is a scalar quantity, representing the dot product of the magnetic field vector B⃗\vec{B} and the area vector A⃗\vec{A}.

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The area vector A⃗\vec{A} is always directed perpendicular (normal) to the surface.

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The angle θ\theta used in the calculation is the angle between the magnetic field B⃗\vec{B} and the normal to the plane of the area.

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Magnetic flux is maximum (Φ=BA\Phi = BA) when the magnetic field is perpendicular to the plane (θ=0∘\theta = 0^\circ).

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Magnetic flux is zero when the magnetic field is parallel to the plane of the area (θ=90∘\theta = 90^\circ).

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The SI unit of magnetic flux is Weber (WbWb), where 1 Wb=1 Tesla⋅meter21\text{ Wb} = 1\text{ Tesla} \cdot \text{meter}^2.

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The dimensional formula for magnetic flux is [ML2T−2A−1][ML^2T^{-2}A^{-1}].

📐Formulae

ΦB=B⃗⋅A⃗\Phi_B = \vec{B} \cdot \vec{A}

ΦB=BAcos⁡θ\Phi_B = BA \cos \theta

Φtotal=NΦB=NBAcos⁡θ\Phi_{total} = N \Phi_B = NBA \cos \theta

Dimensional Formula=[ML2T−2A−1]\text{Dimensional Formula} = [ML^2T^{-2}A^{-1}]

💡Examples

Problem 1:

A square loop of side 10 cm10\text{ cm} is placed in a uniform magnetic field of 0.5 T0.5\text{ T}. If the normal to the loop makes an angle of 60∘60^\circ with the magnetic field, calculate the magnetic flux through the loop.

Solution:

Given: Side s=10 cm=0.1 ms = 10\text{ cm} = 0.1\text{ m}. Area A=s2=(0.1)2=0.01 m2A = s^2 = (0.1)^2 = 0.01\text{ m}^2. B=0.5 TB = 0.5\text{ T} and θ=60∘\theta = 60^\circ. Using Φ=BAcos⁡θ\Phi = BA \cos \theta: Φ=0.5×0.01×cos⁡60∘\Phi = 0.5 \times 0.01 \times \cos 60^\circ Φ=0.005×0.5=0.0025 Wb\Phi = 0.005 \times 0.5 = 0.0025\text{ Wb}

Explanation:

The flux is calculated by taking the component of the magnetic field along the normal to the area. Since cos⁡60∘=0.5\cos 60^\circ = 0.5, the effective flux is half of the maximum possible flux for this orientation.

Problem 2:

Calculate the change in magnetic flux (ΔΦ\Delta \Phi) if the initial flux through a coil is 850 mWb850\text{ mWb} and it is reduced to 320 mWb320\text{ mWb} when the coil is rotated.

Solution:

Initial flux Φ1=850 mWb\Phi_1 = 850\text{ mWb}, Final flux Φ2=320 mWb\Phi_2 = 320\text{ mWb}. Calculation of change: 850−320530\begin{array}{r} 850 \\ -320 \\ \hline 530 \end{array} The change in flux is ΔΦ=530 mWb=0.53 Wb\Delta \Phi = 530\text{ mWb} = 0.53\text{ Wb}.

Explanation:

The change in flux is the absolute difference between the initial and final states, which is essential for determining induced EMF using Faraday's Law.