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

Answer:

$\phi_B = 0.216 T m^2$

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

$L = 4a$

$a = \frac{L}{4} = 0.5 m$

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

$\theta = 30.0^o$

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

$\phi_B = 0.216 T m^2$