Question

Figure *a *shows a circuit consisting of an ideal battery with emf $$\mathscr{E}=6.00\muV$$, a resistance *R*, and a small wire loop of area $5.0cm_{2}$. For the time interval *t *= 10 s to *t *= 20 s, an external magnetic field is set up throughout the loop.The field is uniform, its direction is into 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 $i_{s}=2.0mA$. Find the constant *a *in the equation for the field magnitude.

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Updated on : 2022-09-05

Solution

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From the datum at $t=0$ in Fig. b

we see $0.0015A=V_{battery}/R,$ which implies that the

resistance is

$R=(6.00μV)/(0.0015A)=0.0040Ω$

Now, the value of the current during

$10s<t<20s$ leads us to equate

$(V_{battery}+ε_{induced})/R=0.00050A$

This shows that the induced emf is $ε_{induced}=−4.0μV$. Now we use Faraday's law:

$ε=−dtdΦ_{B} =−AdtdB =−Aa$

Plugging in $ε=−4.0×10_{−6}V$ and $A=5.0×10_{−4}m_{2},$ we

obtain $a=0.0080T/s$

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