Question

An equilateral triangle 8.0 cm on a side is in a 3.5 mT uniform magnetic field. The magnetic flux through the triangle is 6.0 μWb. What is the angle between the magnetic field and an axis perpendicular to the plane of the triangle?

Answer 1

Answer:

43.80°

Explanation:

The side of the equilateral triangle is given as 8 cm

The area of the equilateral triangle $A=\frac{\sqrt{3}}{4}a^2=\frac{1.732\times 8^2}{4}=27.712cm^2=27.712\times 10^{-4}m^2$

Magnetic field $B=3.5mT=3.5\times 10^{-3}T$

Magnetic flux $\Phi =6\mu Wb=6\times 10^{-6}weber$

We know that  $\Phi =BACOS\Theta$

So $6\times 10^{-6}=3\times 10^{-3}\times 27.712\times 10^{-4}cos\Theta$

$cos\Theta =0.7217$

$\Theta =43.80^{\circ}$

Answer 2

The angle between the magnetic field and an axis perpendicular to the plane of the triangle is 51.8 ⁰C.

Area of the equilateral triangle

The area of the equilateral triangle is calculated as follows;

$A = \frac{a^2\sqrt{3} }{4} \\\\A = \frac{(0.08)^2 \times \sqrt{3} }{4} \\\\A = 2.77 \times 10^{-3} \ m^2$

Magnetic flux through the triangle

Ф = BA cosθ

where;

• B is magnetic field strength
• A is the area of the triangle
• θ is the angle between the magnetic field and an axis perpendicular to the plane of the triangle

$cos \theta = \frac{\Phi}{BA} \\\\\theta = cos^{-1} (\frac{\Phi}{BA})\\\\\theta = cos^{-1} (\frac{6 \times 10^{-6} }{3.5 \times 10^{-3} \times 2.77 \times 10^{-3} })\\\\\theta = 51.8 \ ^0C$

Thus, the angle between the magnetic field and an axis perpendicular to the plane of the triangle is 51.8 ⁰C.

Learn more about magnetic flux here: https://brainly.com/question/10736183