Physics: 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?

When the magnetic flux through a single loop of

Ответ:

mpateltx

21.01.2022

COMPLETE QUESTION:

When the magnetic flux through a single loop of wire increases by 30 Tm^2 , an average current of 40 A is induced in the wire. Assuming that the wire has a resistance of 2.5 ohms , (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?

(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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Ответ:

kinggonzales5082

16.02.2022

Physics

What is the direction of the frictional force?

Direction of friction is always opposite to relative motion

Explanation:

Friction always resist the relative motion of two surface so it is always opposite to the direction of relative velocity

What is the direction of the normal force?

It is perpendicular to the contact plane

Explanation:

Normal force is the perpendicular force which is always at 90 degree with the contact surface

How does the frictional force depend on the normal force?

$F_f = \mu F_n$

Explanation:

Friction force is directly proportional to the normal force so we can say it is given as

$F_f = \mu F_n$

Let the x-axis be parallel to the incline and the y-axis rise from the incline at a right angle. By Newton's second law, if there is no acceleration in the y-direction (perpendicular to the plane), what must be the magnitude of the normal force?

$F_n = mg cos\theta$