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Physics · Ch 4 — Electromagnetic Induction and Alternating Current

Magnetic Flux ($\Phi_B$)

4.1.2

Magnetic Flux ($\Phi_B$)

Magnetic flux (ΦB\Phi_B) through an area A in a magnetic field is defined as the number of magnetic field lines passing normally through that area. Formally, for a general (possibly curved, possibly non-uniform-field) surface, it is the surface integral

ΦB=∫B⃗⋅dA⃗(4.1)\Phi_B = \int \vec B\cdot d\vec A \qquad (4.1)

where the integral runs over the whole area A and, at each small element dA⃗d\vec A, the dot product picks out only the component of B⃗\vec B along that element's own outward normal. For the much more common special case used throughout this unit -- a FLAT area A with a UNIFORM field B⃗\vec B making a fixed angle θ\theta with the area's normal -- this reduces to the simple product

ΦB=BAcos⁡θ\Phi_B = BA\cos\theta

which becomes the maximum possible value ΦB=BA\Phi_B=BA when the field is exactly along the normal (θ=0∘\theta=0^{\circ}, since cos⁡0∘=1\cos0^{\circ}=1), and becomes exactly zero when the field lies entirely IN the plane of the area (θ=90∘\theta=90^{\circ}, since cos⁡90∘=0\cos90^{\circ}=0) -- a common source of error is to use the angle between the field and the PLANE of the area instead of the angle between the field and the area's NORMAL, which are always 90∘90^{\circ} apart from each other. …

Figure 4.1Magnetic flux

What this figure shows. Two panels illustrate the definition of magnetic flux through a small area element. Panel (a) shows a general curved surface divided into small area elements dA⃗d\vec A, each with its own outward normal n^\hat n, immersed in a magnetic field B⃗\vec B that need not be uniform or perpendicular to every element -- the flux through the whole surface is the sum (integral) of B⃗⋅dA⃗\vec B\cdot d\vec A over all these elements. Panel (b) shows the simpler special case used for most calculations in this unit: a flat area A with a uniform field B⃗\vec B making a fixed angle θ\theta with the area's normal n^\hat n, for which the flux reduces to the single product $ …

Misc Example 4.1Flux linked with a tilted circular antenna

Worked out. A circular antenna of area 3 m2^2 installed in Madurai has its plane inclined at 47∘47^{\circ} to Earth's magnetic field of magnitude 4.1×10−54.1\times10^{-5} T, and the question asks for the flux linked with it. Because the flux formula uses the angle between the field and the area's NORMAL, not the angle between the field and the plane itself, the working angle is θ=90∘−47∘=43∘\theta = 90^{\circ}-47^{\circ}=43^{\circ}. Substituting into ΦB=BAcos⁡θ\Phi_B=BA\cos\theta gives ΦB=(4.1×10−5)(3)(cos⁡43∘)=(4.1×10−5)(3)(0.7314)≈89.96 μWb\Phi_B = (4.1\times10^{-5})(3)(\cos43^{\circ}) = (4.1\times10^{-5})(3)(0.7314) \approx 89.96\ \mu\text{Wb}, illustrating the common trap of confusing the angle a plane makes with a f …

Misc Example 4.2Flux through a rotating loop at three orientations

Worked out. A circular loop of area 5×10−25\times10^{-2} m2^2 rotates about a diameter perpendicular to a uniform field of 0.2 T, and the flux is required when its plane is (i) normal to the field, (ii) inclined 60∘60^{\circ} to the field, and (iii) parallel to the field. Working each case through ΦB=BAcos⁡θ\Phi_B=BA\cos\theta with the normal-to-field angle: (i) plane normal to field means the loop's own normal is ALONG the field, so θ=0∘\theta=0^{\circ} and ΦB=(0.2)(5×10−2)(1)=1×10−2\Phi_B = (0.2)(5\times10^{-2})(1) = 1\times10^{-2} Wb; (ii) plane inclined 60∘60^{\circ} to the field means the normal is at θ=90∘−60∘=30∘\theta=90^{\circ}-60^{\circ}=30^{\circ}, giving ΦB=(0.2)(5×10−2)(cos⁡30∘)≈0.866×10−2\Phi_B=(0.2)(5\times10^{-2})(\cos30^{\circ}) \approx 0.866\times10^{-2} Wb; (iii) plane parallel to the field means the normal is at θ=90∘\theta=90^{\circ}, so cos⁡90∘=0\cos90^{\circ}=0 and ΦB=0\Phi_B=0. The worked example drives home that flux is maximum when the plane is perpendicular to th …

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