# Electricity Exam Questions

1. a) State the Faraday’s Law. b) Calculate the magnetic flux for the case of a 40 cm radius circular loop. The magnetic field vector makes an angle of 30 degree with the loop’s plane with the magnitude set to 2 Tesla. Provide a sketch. Consider that fact that the loop has a resistance (R) of 2.5 Ohms. Calculate i) the induced current and ii) indicate (on your drawing) its direction at t=1 sec if the magnetic field is changing in time as follows: $B = B0 +B1*t$,  $B1 =0.3$ T/s, $B0 = 2T$

2. A circular loop has a radius of R=2.5cm and carries 0.5A current (IL) flowing in the counter-clockwise direction. The center of the loop is a distance $d1=10$ cm above a long straight wire carrying current $I1=2A$ f1owing to the right. In addition, another long wire with current $I2=3$ A flowing in the downward direction is placed at a distance $d2=15$ cm to the left from the center. a) What is the resulting (total) magnetic field (magnitude and direction) at the center of V the loop? b) Calculate magnitudes and indicate directions for forces acting on short (d(l)=1mm) segments of the loop at point a. c) Calculate the same for point b.

3. In the circuit shown below $E=9V$ , $R1=15$ Ohms, and $L=0.1$H. Switch S1 is closed at t=0 s. Just after the switch is closed

Part I.

a) What is the potential difference ($V_{bc}=V_b-V_c$) across the inductance?

b) Potentials at points a and b are in the following relationships:

i)Point a is at higher potential than b; ii) Point b is at higher potential than a iii) The potentials are equal iv)Not enough information.

Part II.

c) What is the maximum value for current through Ri at sufficiently longer times (i.e. very long time after the switch was closed)? .

d) How much time (t60%) is needed to reach 60% of that maximum value?

e) What is the rate of current change when it reached the target value (i.e. 60% of the maximum)? a)

The induced voltage in a circuit that is located into a changing magnetic field is

equal with the speed of variation of the magnetic flux through the surface of the circuit,

taken with changed sign (-sign).

b)

Surface area

$S = pi*r^2 =pi*0.4^2 =0.503 m^2$

flux

$phi = B*S*cos(90-30) =2*0.503*cos(60) =0.503 Wb$

see first figure attached for the sketch.

c)

$B= B0+B1*t$

If the field is perpendicular to the surface

$phi_0 = (B0+B1*t)*S$

$V_{ind}0 = – d(phi_0)/d t = – B1*S= 0.3*0.503 = 0.1509 V$

$I_{ind}0 = V_{ind}0/R =0.1509/2.5 =0.06036 A =60.36 mA$

If the field is making a 30 degree angle with loop plane

$phi = phi_0*cos(90-30)$

$V_{ind} = V_{ind}0*cos(90-30) =0.07545 V$

$I_{ind}0 = V_{ind}/R =0.7545/2.5 =0.03018 A =30.18 mA$

(the induced current is opposing the variation of the magnetic field, see attached picture)

2.

a)

$mu =4*pi*10^{-7} (H/m)$

field of loop alone

$B(loop) = mu*I_L/(2*R) =1.256*10^{-5} T$ out of page

field of current $I_1$ is

$B(I_1) = mu*I_1/(2*pi*d_1) =4*10^{-6} T$ out of page

field of current $I_2$ is

$B(I2) = mu*I_2/(2*pi*d2) =4*10^{-6} T$ out of page

Total field $B = (12.56+4+4)*10^{-6} =20.56*10^{-6} T$ out of page

b)

$B(I_1) = mu*I_1/(2*pi*(d_1-R)) =5.33*10^{-6} T$

$F(I_1) = (B(I_1) x d_l)*I_L =5.33*10^{-6}*0.001*0.5 =2.67*10^{-9} N$

directed from a to the center of loop

$B(I_2)= 4*10^6 T$

$F(I_2) =(B(I_2) x d_l)*I_L =4*10^{-6}*0.001*0.5 =2*10^{-9} N$

directed also from point a to the center of loop

$F = F(I_1)+F(I_2) = 4.67*10^{-9} N$

directed also from point a to the center of loop.

c)

$B(I_1) =4*10^{-6} T$

$F(I_1) = (B(I_1) x d_l)*I_L = 2*10^{-9} N$

directed from point b to the center of loop

$B(I_2) = mu*I_2/(2*pi*(d_2+R)) =3.42*10^{-6} T$

$F(I_2) = (B(I_2) x d_l)*I_L = 1.71*10^{-9} N$

directed from point b to the center of loop

$F = (1.71+2)*10^{-3} = 3.71*10^{-9} N$

directed from point b to the center of loop

3.

Part I)

Just after the switch is closed

a) $V_{bc} = V_b- V_c = E = +9 V$

(the induced voltage in the coil opposes the increasing of current in the circuit)

b)

At the first moment there is no current in the circuit. Hence

$V_a = V_b > V_c$

Part II)

c) at very long time coil is acting like a short circuit.

$I(R_1) = E/R_1 = 9/15 =0.6 A$

d)

$I = I_{max}(1-exp(-t/tau))$

$tau =L/R = 0.1/15 =6.67 ms$

$I/I_{max} = 1-exp(-t/0.00667)$

$0.6 =1-exp(-t/0.00667)$

$0.4 = exp(-t/0.00667)$

$t/0.00667 =0.916$

$t =0.00611 =6.11 ms$

e)

$d I/d t = I_{max}/tau*exp(-t/tau)$

at $t=6.11 ms$

$d I/d t = (0.6/0.00667)*exp(-6.11/6.667) = 36 (A/s)$