A single-turn plane loop of wire with a cross-sectional area 200 cm2 is perpendicular to a magnetic field that increases uniformly from 0.200 T to 2.800 T in 2.20 seconds. What …

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asked Dec 4, 2020 225k views

1 Answer

Marcin D · Answer 1 · 2020-12-10T06:26:34+0000

Answer:

Induced current,
$I=2.87* 10^(-3)\ A$

Step-by-step explanation:

Given that,

Area of cross section of the wire,
$A=200\ cm^2=0.02\ m^2$

Time, t = 2.2 s

Initial magnetic field,
$B_i=0.2\ T$

Final magnetic field,
$B_f=2.8\ T$

Resistance of the coil, R = 8 ohms

The expression for the induced emf is given by :

$\epsilon=-(d\phi)/(dt)$

$\phi$ = magnetic flux

$\epsilon=-(d(BA))/(dt)$

$\epsilon=A(d(B))/(dt)$

$\epsilon=A(B_f-B_i)/(t)$

$\epsilon=0.02* (2.8-0.2)/(2.2)$

$\epsilon=-0.023\ volts$

So, the induced emf in the loop is 0.023 volts. The induced current can be calculated using Ohm's law as :

$\epsilon= IR$

$I=(\epsilon)/(R)$

I=(0.023)/(8)

$I=2.87* 10^(-3)\ A$

So, the magnitude of the induced current in the loop of wire is
$2.87* 10^(-3)\ A$ . Hence, this is the required solution.