When in use, a lamp is designed to draw

Physics

1 answer:

Ganezh [65]2 years ago

8 0

Answer:

R=∆V/I=12/0.6=20ohm....

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Answer:

Approximately 18 volts when the magnetic field strength increases from $\rm 20\; mT$ to $\rm 80\;mT$ at a constant rate.

Explanation:

By the Faraday's Law of Induction, the EMF $\epsilon$ that a changing magnetic flux induces in a coil is:

$\displaystyle \epsilon = N \cdot \frac{d\phi}{dt}$ ,

where

$N$ is the number of turns in the coil, and
$\displaystyle \frac{d\phi}{dt}$ is the rate of change in magnetic flux through this coil.

However, for a coil the magnetic flux $\phi$ is equal to

$\phi = B \cdot A\cdot \cos{\theta}$ ,

where

$B$ is the magnetic field strength at the coil, and
$A\cdot \cos{\theta}$ is the area of the coil perpendicular to the magnetic field.

For this coil, the magnetic field is perpendicular to coil, so $\theta = 0$ and $A\cdot \cos{\theta} = A$ . The area of this circular coil is equal to $\pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}$ .

$A\cdot \cos{\theta} = A$ doesn't change, so the rate of change in the magnetic flux $\phi$ through the coil depends only on the rate of change in the magnetic field strength $B$ . The size of the magnetic field at the instant that $B = \rm 50\; mT$ will not matter as long as the rate of change in $B$ is constant.

$\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}$ .