Physics, 04.08.2020 14:01 maddbroms

In a physics laboratory experiment, a coil with 250 turns enclosing an area of 14 cm2 is rotated in a time interval of 0.030 s from a position where its plane is perpendicular to the earth's magnetic field to a position where its plane is parallel to the field. The earth's magnetic field at the lab location is 5.0×10^−5 T. Required:a. What is the total magnetic flux through the coil before it is rotated? After it is rotated? b. What is the average emf induced in the coil?

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Answer from: mckenzieg

Explanation:

Consider a loop of wire, which has an area of $A=14 \mathrm{cm}^{2}$ and N=250 turns, it is initially placed perpendicularly in the earth magnetic field. Then it is rotated from this position to a position where its plane is parallel to the field as shown in the following figure in $\Delta t=0.030$ s. Given that the earth's magnetic field at the position of the loop is $B=5.0 \times 10^{-5} \mathrm{T}$ , the flux through the loop before it is rotated is,

$\Phi_{B, i} &=B A \cos \left(\phi_{i}\right)=B A \cos \left(0^{\circ}\right$

$=\left(5.0 \times 10^{-5} \mathrm{T}\right)\left(14 \times 10^{-4} \mathrm{m}^{2}\right)(1)$

$=7.0 \times 10^{-8} \mathrm{Wb}$

$\quad\left[\Phi_{B, i}=7.0 \times 10^{-8} \mathrm{Wb}\right$

after it is rotated, the angle between the area and the magnetic field is $\phi=90^{\circ}$ thus,

$\Phi_{B, f}=B A \cos \left(\phi_{f}\right)=B A \cos \left(90^{\circ}\right)=0$

$\qquad \Phi_{B, f}=0$

(b) The average magnitude of the emf induced in the coil equals the change in the flux divided by the time of this change, and multiplied by the number of turns, that is,

${\left|\mathcal{E}_{\mathrm{av}}\right|=N\left|\frac{\Phi_{B, f}-\Phi_{B, i}}{\Delta t}\right|}{=} & \frac{1.40 \times 10^{-5} \mathrm{Wb}}{0.030 \mathrm{s}}$

$& 3.6 \times 10^{-4} \mathrm{V}=0.36 \mathrm{mV}$

$\mathbb{E}=0.36 \mathrm{mV}$

Answer from: Quest

Well the magnitude of its acceleration related to that car starts with the oil change like when you have to change the oil in your car you have to make sure that your precious car is starting really propley and very well

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

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