Question

A 2.0 m conductor is formed into a square and placed in the horizontal xy-plane. A magnetic field is oriented 30.0° above the horizontal with a strength of 1.0 T. What is the magnetic flux through the conductor?

Answer 1

`\phi_B = 0.216 T m^2`

Step-by-step explanation:

As we know that the length of the conductor is given as

`L = 2 m`

now if it is converted into a square then we have

`L = 4a`

`a = (L)/(4) = 0.5 m`

now the are of the loop will be

`A = a^2 = 0.5(0.5) = 0.25 m^2`

now the magnetic flux is defined as

`\phi_B = BAcos\theta`

here we know

B = 1.0 T

`\theta = 30.0^o`

`\phi_B = (1.0 T)(0.25 m^2)(cos30)`

`\phi_B = 0.216 T m^2`