What approximately is the percent uncertainty for the measurement 5.2

Each of the two coils has 320 turns. The average radius of the coil is 6 cm. The distance from the center of one coil to the ele

Temka [501]

Answer:

Magnetic field, $B=2.39\times 10^{-3}\ T$

Explanation:

It is given that,

Number of turns, N = 320

Radius of the coil, r = 6 cm = 0.06 m

The distance from the center of one coil to the electron beam is 3 cm, x = 3 cm = 0.03 m

Current flowing through the coils, I = 0.5 A

We need to find the magnitude of the magnetic field at a location on the axis of the coils, midway between the coils. The magnetic field midway between the coils is given by :

$B=\dfrac{\mu_oINr^2}{(x^2+r^2)^{3/2}}$

$B=\dfrac{4\pi \times 10^{-7}\times 0.5\times 320\times (0.06)^2}{(0.03^2+0.06^2)^{3/2}}$

B = 0.00239 T

or

$B=2.39\times 10^{-3}\ T$

So, the magnitude of the magnetic field at a location on the axis of the coils, midway between the coils is $2.39\times 10^{-3}\ T$ . Hence, this is the required solution.

7 0

4 years ago

A 1.2 nf parallel-plate capacitor has an air gap between its plates. Its capacitance increases by 3.0 nf when the gap is filled

Damm [24]

Answer:

2.5

Explanation:

The capacitance of a parallel-plate capacitor filled with dielectric is given by

$C=kC_0$

where

k is the dielectric constant

$C_0$ is the capacitance of the capacitor without dielectric

In this problem,

$C_0=1.2 nF$ is the capacitance of the capacitor in air

$C=3.0 nF$ is the capacitance with the dielectric inserted

Solving the equation for k, we find

$k=\frac{C}{C_0}=\frac{3.0 nF}{1.2 nF}=2.5$

3 0

3 years ago

Read 2 more answers

A rock is tossed straight up from the ground with a speed of 21 m/s . When it returns, it falls into a hole 10 m deep.a.) What i

Arte-miy333 [17]

(a) 25.2 m/s

Let's take the initial vertical position of the rock as "zero" (reference height).

According to the law of conservation of energy, the speed of the rock as it reaches again the position "zero" after being thrown upwards is equal to the initial speed of the rock, 21 m/s (in fact, if there is no air resistance, no energy can be lost during the motion; and since the kinetic energy depends only on the speed of the rock:

$K=\frac{1}{2}mv^2$

and the gravitational potential energy of the rock has not changed, since the rock has returned into its initial position, it means that the speed of the rock should be the same)

This means that we can only analyze the final part of the motion, the one in which the rock falls into the 10 m hole. Since it is a free fall motion, we can find the final speed by using

$v^2 = u^2 + 2gd$