Question #ed14d

Question

Question #ed14d

1 Answer

Impact of this question

2292 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-18T17:09:01

Given

$B \to Magnetic field = 0.230 T in XZ direction$ $B->"Magnetic field"=0.230T" in XZ direction"$

$A \to Area = π \times {(0.065)}^{2} m^{2} in XY plane$ $A->"Area"=pixx(0.065)^2m^2" in XY plane"$

The angle subtended by the magnetic field in the XZ direcion with the direction of area vector i.e normal to the plane of the circular area is $θ = 45^{\circ}$ $theta=45^@$

So the mag. flux through the area $ϕ = B \cdot A \cdot cos θ$ $phi=B*A*costheta$

$= 0.23 \cdot π \cdot {(0.065)}^{2} \cdot {cos 45}^{\circ} W b$ $=0.23*pi*(0.065)^2*cos45^@Wb$

$= 0.002 W b$ $=0.002Wb$

Science

Math

Humanities

... and beyond

Question #ed14d

1 Answer

Related questions

Impact of this question