Calculate the magnetic flux through the trian | vedclass

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Question

(B) The magnetic flux $\phi$ is given by the formula $\phi = \vec{B} \cdot \vec{A} = BA \cos 	heta$.Here,the magnetic field $B = 2 	ext{ T}$ is perp

Vedclass · Accepted Answer

$2 	imes 10^{-4} 	ext{ Wb}$

Vedclass · Answer

(B) The magnetic flux $\phi$ is given by the formula $\phi = \vec{B} \cdot \vec{A} = BA \cos 	heta$.Here,the magnetic field $B = 2 	ext{ T}$ is perpendicular to the area,so the angle $	heta = 0^{\circ}$ and $\cos 0^{\circ} = 1$.The area $A$ of the right-angled triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Given base $= 2 	ext{ cm} = 2 	imes 10^{-2} 	ext{ m}$ and height $= 1 	ext{ cm} = 1 	imes 10^{-2} 	ext{ m}$.$A = \frac{1}{2} 	imes (2 	imes 10^{-2} 	ext{ m}) 	imes (1 	imes 10^{-2} 	ext{ m}) = 1 	imes 10^{-4} 	ext{ m}^2$.Now,calculating the flux: $\phi = 2 	ext{ T} 	imes (1 	imes 10^{-4} 	ext{ m}^2) 	imes 1 = 2 	imes 10^{-4} 	ext{ Wb}$.

Calculate the magnetic flux through the triangular loop shown in the figure. $A$ uniform magnetic field of strength $2 \text{ T}$ points perpendicularly into the plane of the triangle.

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