# Answer in Electric Circuits for farah #100498

A linked magnetics flux in a coil can be calculated according to formula

(1) “Phi=oint overline B cdot doverline s=overline B cdot overline S= B cdot S cdot cos(widehat{BS})”, as “overline B=const” and “S” is flat.

In our case we have “B=0.03 T, S=pi frac{D^2}{4}N , D=0.1 m (10cm),N=50” turns. When determining the total area of the coil, of course, it is assumed that all the turns are wound in the same direction and their areas “S=pi frac{D^2}{4}” should be summed. “S=3.14 frac{0.1^2}{4} 50=0.39 m^2”

If a linked magnetics flux change with time the induced e.m.f. one should determine according Faraday’s law of electromagnetic induction “V=-frac{partialPhi}{partial t}”. In our case “Phi” depends on the time due to flipped the coil. What physical quantity changes in this process? Note that the surface appearing in formula (1) is considered to have a direction coinciding with its normal. Therefore when the coil is perpendicular to the direction of the magnetic flux the vectors “overline B” and “overline S” are collinear and their scalar product coincides with the product of their modules. If these vectors form an angle, the formula for the magnetic flux corresponds to the scalar product of these vectors. Before flipping the linked magnetics flux is “Phi_0= B cdot S”. After flipping it is “Phi_1= B cdot S cdot cos60degree”. The difference is “delta Phi=B cdot S cdot (cos60degree -1)”.

Then e.m.f. is “V=-frac{delta Phi}{delta t}=frac{B cdot S cdot ( 1-cos60degree)}{delta t}=0.03Tcdot 0.39m^2 frac{1-0.5}{30cdot 10^{-3}s}=0.196 V”

If the self induction of the coil is negligible the induced current can be determine versis the Ohm’s law: “I=V/R=0.196/35=5.61 mA” .

**Answer: The induced e.m.f. is **“0.196 V”** . The induced current flows is **“5.61 mA”**.**