Figure (a) shows a circuit consisting of an ideal battery with emf E = 6.31 mu V

Question

Figure (a) shows a circuit consisting of an ideal battery with emf E = 6.31 mu V, a resistance R, and a small wire loop of area 5.6 cm^2. For the time interval t = 40 s to t = 80 s, an external magnetic field is set up throughout the loop. The field is uniform, its direction is in the page in Figure (a), and the field magnitude is given by B = at, where B is in teslas, a is a constant, and t is in seconds. Figure (b) gives the current i in the circuit before, during, and after the external field is set up. The vertical axis scale is set by i_s = 3.4 mA, and the horizontal axis scale is set by t_s = 120 s. Find a.

Explanation / Answer

We have B = at, 40s<t<80s and 0 otherwise.

Area of the loop is A = 5.6cm2 = 5.6X10-4m2

So flux throught the loop during for 40s<t<80s is,

= B.A =BA = atA

emf induced in the coil is;

E = -d/dt = -d(atA)/dt = -(aA)

So the current induced in the coil is,

I0 = E/R = -(aA)/R ----- (1), this current will be in a direction opposite to that caused by the battery in the loop.

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Current through the coil for 0<t<40 is I. I can be found using the current scale in figure (b). In figure is = 3.4mA. And current I is given by the solid line for 0<t<40s. Using the scale of current;

I = [(3.4mA)/4]X3 = 2.55mA, for 0<t<40 ----------- (2)

for  0<t<40 , voltage in the circuit is the voltage of battery V0 = 6.31X10-6V

So R = V0/I = (6.31X10-6V)/(2.55mA) = 2.474X10-3

Now current through the coil or the circuit for 40s<t<80s is Inet = [(3.4mA)/4]X1 = 0.85mA

And Inet = I + I0, for 40s<t<80s

or 0.85mA = 2.55mA -(aA)/R (putting values of I0 and I from equation (1) and (2))

or 0.85mA = 2.55mA - [a(5.6X10-4m2)]/[2.474X10-3]

or 0.85mA = 2.55mA - a(2.263X10-1)

or a = 7.51X10-3T/s .... is the required value of a.

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This concludes the answers. Check the answer and let me know if it's correct. If you need anymore clarification or correction I will be happy to oblige.....

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