A square of side x , m lies in the x-y plane | vedclass

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Question

(A) The area vector $\vec A$ of a square of side $x$ lying in the $x-y$ plane is given by $\vec A = x^2 \hat k \, m^2$.The magnetic field is given as

Vedclass · Accepted Answer

$5B_0x^2 \, Wb$

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(A) The area vector $\vec A$ of a square of side $x$ lying in the $x-y$ plane is given by $\vec A = x^2 \hat k \, m^2$.The magnetic field is given as $\vec B = B_0 (3\hat i + 4\hat j + 5\hat k) \, T$.The magnetic flux $\phi$ is defined as the dot product of the magnetic field vector and the area vector: $\phi = \vec B \cdot \vec A$.Substituting the values: $\phi = [B_0 (3\hat i + 4\hat j + 5\hat k)] \cdot [x^2 \hat k]$.Using the dot product properties $(\hat i \cdot \hat k = 0, \hat j \cdot \hat k = 0, \hat k \cdot \hat k = 1)$:$\phi = B_0 x^2 (3 \cdot 0 + 4 \cdot 0 + 5 \cdot 1) = 5 B_0 x^2 \, Wb$.

$A$ square of side $x \, m$ lies in the $x-y$ plane in a region where the magnetic field is given by $\vec B = B_0 (3\hat i + 4\hat j + 5\hat k ) \, T$,where $B_0$ is a constant. The magnitude of the magnetic flux passing through the square is:

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