# Magnetic Flux and Faraday’s Law

A square loop of Cu wire 4.00 cm on an edge is lying on a horizontal table. An electromagnet above the square loop is turned on to produce a magnetic field. The magnetic field, directed downward toward the table and at an angle of 60 degrees from the surface of the table, goes through the square loop of Cu wire. (Copper wire for this problem: radium 1.02 mm and resistivity of copper is $1.70e{-8}$ ohms*m)

Find:

a) the induced emf in the loop if the magnetic field due to the electromagnet increases from 0.00T to 0.500T in 0.200s

b) what is the current in the square loop?

c) what is the power loos in the loop?

a) Surface of loop is $S =4*4 =16 cm^2 =16*10^{-4} m^2$

Perpendicular component of B to surface of loop is $B_n = B*sin(\alpha)$,

$\alpha =60 degree$

flux of B through surface is $\Phi = B_n*S = B*S*sin(60)$

Induced voltage is

$V_{ind} =-d(\phi)/d t = 0.5*16*10^{-4}*sin(60)/0.2 =3.46*10^{-3} V =3.46 mV$

b) length of loop $L =4*4 =16 cm$

Resistance of loop $R = \rho*L/S =1.7*10^{-8}*0.16/(\pi*0.00102^2) =8.32*10^{-4} ohm$

$I = V_{ind}/R = 3.46*10^{-3}/8.32*10^{-4} =4.158 A$

c) Power loss is $P = U*I =0.0144 W$