Q.The magnetic flux linked with a coil of N turns of area of cross-section A held with its plane parallel to the field B is

Magnetic flux φ = B·A·cos θ. The plane of the loop is perpendicular to B, so B is along the normal (θ = 0) and φ = BA = 0·5 × 1 = 0·5 Wb. Option (a).

Step 1 — area of the square loop: A = (1 m)² = 1 m².

Step 2 — orientation: 'plane of the loop perpendicular to B' means the field is along the area's normal, so the angle between B and the normal is 0° and cos θ = 1.

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Magnetic flux depends on the angle between the field and the AREA VECTOR (normal to the coil), not the plane itself.

$\Phi = NAB\cos\theta$, where $\theta$ is the angle between $\vec B$ and the normal $\hat n$ to the coil's plane. If the coil's plane is parallel to $\vec B$, t …

Tesla is the unit of magnetic flux density (B), weber is the unit of magnetic flux (Φ = B·A), so weber equals tesla times square metre.

Magnetic flux $\Phi$ through a surface of area $A$ in a uniform field of magnetic flux density $B$ is:

$\Phi = B \cdot A$

Here $B$ is measured in tesla (T), the SI unit of magnetic flux density, and $\Phi$ is measured in weber (Wb), the SI unit of magnetic flux. Since $\Phi = BA$, dimensionally: …

Magnetic flux Φ = B·A; working out the dimension of B from F = qvB and multiplying by area gives [ML²T⁻²A⁻¹].

From F = qvB, the dimension of B = [F]/([q][v]) = (MLT⁻²)/((AT)(LT⁻¹)) = MT⁻²A⁻¹.

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