Ask question

Ask question

All categories

3 years ago

13

A metal rod of length (L) moves with velocity (v), perpendicular to its length, in a magnetic field B, which is perpendicular to

both the rod and its velocity. If the length of the rod is doubled, what happens to the electric field in the rod

1 answer:

Alborosie3 years ago

6 0

Explanation:

If a metal rod of length L moves with velocity v is moving perpendicular to its length, in a magnetic field B, the induced emf is given by :

$\epsilon=Blv$

The electric field in the conductor is given by :

$E=\dfrac{\epsilon}{l}\\\\E=\dfrac{Blv}{l}\\\\E=Bv$

It is clear that the electric field is independent of the length of the rod. If the length of the rod is doubled, the electric field in the rod remains the same.

You might be interested in

The bright yellow light emitted by a sodium vapor lamp consists of two emission lines at 589.0 and 589.6 nm. What are the freque

stiv31 [10]

Answer:

Explanation:

Given

wavelength of emissions are

$\lambda _1=589 nm$

$\lambda _2=589.6 nm$

Energy is given by

$E=\frac{hc}{\lambda }$

where h=Planck's constant

x=velocity of Light

$\lambda$ =wavelength of emission

$E_1=\frac{6.626\times 10^{-34}\times 3\times 10^8}{589\times 10^{-9}}$

$E_1=3.374\times 10^{-19} J$

$E_1 in kJ/mol$

$E_1=203.2 kJ/mol$

frequency corresponding to this emission

$\nu =\frac{c}{\lambda }$

$\nu _1=\frac{3\times 10^8}{589\times 10^{-9}}$

$\nu _1=5.09\times 10^{14} Hz$

Energy corresponding to $\lambda _2$

$E_2=\frac{6.626\times 10^{-34}\times 3\times 10^8}{589.6\times 10^{-9}}$

$E_2=3.371\times 10^{-19} J$

$E_2=203.02 kJ/mol$

frequency corresponding to this emission

$\nu =\frac{c}{\lambda }$

$\nu _1=\frac{3\times 10^8}{589.6\times 10^{-9}}$

$\nu _1=5.088\times 10^{14} Hz$

6 0

3 years ago

A ball rolls off a table with a horizontal velocity of 3 m/s. If it takes 0.3 seconds for the ball to reach the floor, how high

LekaFEV [45]

The height of the table above the ground is 0.45 m.

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

Horizontal velocity (u) = 3 m/s
Time (t) = 0.3 s
Acceleration due to gravity (g) = 10 m/s²
Height (h) =?

<h3>How to determine the height </h3>

The height of the table can be obtained by using the following formula:

h = ½gt²

h = ½ × 10 × 0.3²

h = 5 × 0.09

h = 0.45 m

Thus, the height of the table is 0.45 m

Learn more about motion under gravity:

brainly.com/question/26275209

6 0

2 years ago

The graph shows the federal budget from 1980 to 2010. In which period did the federal budget show the greatest deficit?

kodGreya [7K]

Answer:

2000 to 2010

Explanation:

sorry I'm just guessing

8 0

2 years ago

A python can detect thermal radiation from objects that differ in temperature from their environment as long as the received int

yanalaym [24]

Answer:

10.52 m

Explanation:

The power radiated by a body is given by

P = σεAT⁴ where ε = emissivity = 0.97, T = temperature = 30 C + 273 = 303 K, A = surface area of human body = 1.8 m², σ = 5.67 × 10⁻⁴ W/m²K⁴

P = σεAT⁴ = 5.67 × 10⁻⁸ W/m²K⁴ × 0.97 × 1.8 m² × (303)⁴ = 834.45 W

This is the power radiated by the human body.

The intensity I = P/A where A = 4πr² where r = distance from human body.

I = P/4πr²

r = (√P/πI)/2

If the python is able to detect an intensity of 0.60 W/m², with a power of 834.45 W emitted by the human body, the maximum distance r, is thus

r = (√P/πI)/2 = (√834.45/0.60π)/2 = 21.04/2 = 10.52 m

So, the maximum distance at which a python could detect your presence is 10.52 m.

3 0

3 years ago

A proton (mass m = 1.67 × 10-27 kg) is being accelerated along a straight line at 2.50 × 1012 m/s2 in a machine. If the proton h

Veronika [31]

Answer:

(A) Speed will be $44.18\times 10^4m/sec$

(b) Change in kinetic energy = $1560\times 10^{-19}$

Explanation:

We have given mass of proton $m=1.67\times 10^{-27}kg$

Acceleration of the proton $a=2.50\times 10^{12}m/sec^2$

Initial velocity u = $1.60\times 10^4$ m/sec

Distance traveled by proton s = 3.90 cm = 0.039 m

(a) From third equation of motion we know that

$v^2=u^2+2as$

$v^2=(1.60\times 10^4)^2+2\times 2.5\times 10^{12}\times 0.039$

$v=44.18\times 10^4m/sec$

(b) Initial kinetic energy $KE_I=\frac{1}{2}mv^2=\frac{1}{2}\times 1.67\times 10^{-27}\times (1.6\times 10^4)^2$

Final kinetic energy $KE_F=\frac{1}{2}mv^2=\frac{1}{2}\times 1.67\times 10^{-27}\times (44.18\times 10^4)^2$

So change in kinetic energy $\Delta KE=KE_F-KE_I=\frac{1}{2}\times 1.6\times 10^{-27}\times 10^8\times (44.18^2-1.6^2)=1560\times 10^{-19}J$

6 0

3 years ago