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s2008m [1.1K]
3 years ago
9

When the magnetic flux through a single loop of wire increases by , an average current of 40 A is induced in the wire. Assuming

that the wire has a resistance of , (a) over what period of time did the flux increase? (b) If the current had been only 20 A, how long would the flux increase have taken?
Physics
1 answer:
Zielflug [23.3K]3 years ago
7 0

COMPLETE QUESTION:

<em>When the magnetic flux through a single loop of wire increases by </em>30 Tm^2<em> , an average current of 40 A is induced in the wire. Assuming that the wire has a resistance of </em><em>2.5 ohms </em><em>, (a) over what period of time did the flux increase? (b) If the current had been only 20 A, how long would the flux increase have taken?</em>

Answer:

(a). The time period is 0.3s.

(b). The time period is 0.6s.

Explanation:

Faraday's law says that for one loop of wire the emf \varepsilon is

(1). \: \: \varepsilon = \dfrac{\Delta \Phi_B}{\Delta t }

and since from Ohm's law

\varepsilon  = IR,

then equation (1) becomes

(2). \: \:IR= \dfrac{\Delta \Phi_B}{\Delta t }.

(a).

We are told that the change in magnetic flux is \Phi_B = 30Tm^2,  the current induced is I = 40A, and the resistance of the wire is R = 2.5\Omega; therefore, equation (2) gives

(40A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },

which we solve for \Delta t to get:

\Delta t = \dfrac{30Tm^2}{(40A)(2.5\Omega)},

\boxed{\Delta t = 0.3s},

which is the period of time over which the magnetic flux increased.

(b).

Now, if the current had been I =20A, then equation (2) would give

(20A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },

\Delta t = \dfrac{30Tm^2}{(20A)(2.5\Omega)},

\boxed{\Delta t = 0.6 s\\}

which is a longer time interval than what we got in part a, which is understandable because in part a the rate of change of flux \dfrac{\Delta \Phi_B}{\Delta t} is greater than in part b, and therefore , the current in (a) is greater than in (b).

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