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3 years ago

What is the relationship between electric and magnetic fields?

2 answers:

4 0

Are there any answer choices but if its not this is the best answer i can give you, <span>The magnetic field will be perpendicular to the electric field and vice versa An electric field is the area which surrounds an electric charge within which it is capable of exerting a perceptible force on another electric charge. A magnetic field is the area of force surrounding a magnetic pole, or a current flowing through a conductor, in which there is a magnetic flux. A magnetic field can be produced when an electric current is passed through an electric circuit wound in a helix or solenoid. The relationship that exists between an electric field and a magnetic field is one of electromagnetic interaction as a consequence of associating elementary particles. The electrostatic force between charged particles is an example of this relationship.

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geniusboy [140]3 years ago

3 0

Moving point charges, such as electrons, produce complicated but well known magnetic fields that depend on the charge, velocity, and acceleration of the particles. Magnetic field lines form in concentric circles around a cylindrical current-carrying conductor, such as a length of

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An electron that has a velocity with x component 1.6 × 106 m/s and y component 2.4 × 106 m/s moves through a uniform magnetic fi

Sergio039 [100]

Answer:

(a) 5.056 x 10^-14 N

(b) 5.056 x 10^-14 N

Explanation:

X component of velocity of electron is 1.6 × 10^6 m/s

Y component of velocity of electron is 2.4 × 10^6 m/s

X component of magnetic field is 0.025 T

Y component of magnetic field is -0.16 T

charge on electron, q = - 1.6 x 10^-19 C

Write the velocity and magnetic field in the vector forms.

$\overrightarrow{v}=1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j}$

$\overrightarrow{B}=0.025\widehat{i}-0.16\widehat{j}$

The force on the charge particle when it is moving in the magnetic field is given by

$\overrightarrow{F}=q\left ( \overrightarrow{v}\times \overrightarrow{B} \right )$

(a) Force on electron is given by

$\overrightarrow{F}=-1.6\times 10^{-19}\left ( 1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j} \right )\times \left ( 0.025\widehat{i}-0.16\widehat{j} \right )$

$\overrightarrow{F}=5.056\times 10^{-14}\widehat{k}$

Magnitude of force is 5.056 x 10^-14 N.

(b) Force on a proton is given by

$\overrightarrow{F}=1.6\times 10^{-19}\left ( 1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j} \right )\times \left ( 0.025\widehat{i}-0.16\widehat{j} \right )$

$\overrightarrow{F}=-5.056\times 10^{-14}\widehat{k}$

Magnitude of force is 5.056 x 10^-14 N.

Thus, the magnitude of force remains same but the direction of force is opposite to each other.

Explanation:

4 0

3 years ago

A hollow cylinder is given a velocity of 5.3 m/s and rolls up an incline to a height of 2.87 m. If a hollow sphere of the same m

Valentin [98]

Answer:

E = 1/2 M V^2 + 1/2 I ω^2 = 1/2 M V^2 + 1/2 I V^2 / R^2

E = 1/2 M V^2 (1 + I / (M R^2))

For a cylinder I = M R^2

For a sphere I = 2/3 M R^2

E(cylinder) = 1 + 1 = 2 omitting the 1/2 M V^2

E(sphere) = 1 + 2/3 = 1.67

E(cylinder) / E(sphere) = 2 / 1.67 = 1.2

The cylinder initially has 1.20 the energy of the sphere

The PE attained is proportional to the initial KE

H(sphere) = 2.87 / 1/2 = 2.40 m since it has less initial KE

5 0

2 years ago

Determine the density of an object that has a mass of 149.8 g and displaces 12.1 mL of water when placed in a graduated cylinder

Crazy boy [7]

Answer:

"12.4 g/mL" would be the right solution.

Explanation:

The given values are:

Mass of the object,

= 149.8 g

Volume of the object,

= 12.1 mL

Now,

The Density will be:

⇒ $Density=\frac{Mass}{Volume}$

On substituting the given values in the above formula, we get

⇒ $=\frac{149.8}{12.1}$

⇒ $=12.4 \ g/mL$

6 0

3 years ago

At a local swimming pool, the diving board is elevated h = 9.5 m above the pool's surface and overhangs the pool edge by L = 2 m

eimsori [14]

Answer:

1) The time it takes the diver to move off the end of the diving board to the pool surface, $t_w$ , is approximately 1.392 seconds

2) The horizontal distance from the edge of the pool to where the diver enters the water, $d_w$ , is approximately 5.76 meters

Explanation:

1) The given parameters are;

The height of the diving board above the pool's surface, h = 9.5 m

The length by which the diving board over hangs the pool L = 2 m

The speed with which the diver runs horizontally along the diving board, v₀ = 2.7 m/s

Taking $t_w$ = The time it takes the diver to move off the end of the diving board to the pool surface

Therefore, we have from the equation of free fall;

h = 1/2 × g × $t_w$ ²

Where;

g = The acceleration due to gravity = 9.81 m/s²

Substituting the values, gives;

9.5 = 1/2 × 9.81 × $t_w$ ²

$t_w$ = √(9.5/(1/2 × 9.81)) ≈ 1.392 s

The time it takes the diver to move off the end of the diving board to the pool surface = $t_w$ ≈ 1.392 s

2) The horizontal distance, $d_w$ , in meters from the edge of the pool to where the diver enters the water is given as follows;

$d_w$ = L + v₀ × $t_w$ = 2 + 2.7× 1.392 ≈ 5.76 m

∴ The horizontal distance from the edge of the pool to where the diver enters the water ≈ 5.76 meters.

7 0

3 years ago

An unstable atomic nucleus of mass 1.58 10-26 kg initially at rest disintegrates into three particles. One of the particles, of

lara31 [8.8K]

Answer:

a The velocity of the third particle is $v_3 = (-15.629*10^{6}i -14.45*10^{6}j)\ m/s$

b The total kinetic energy increase $E = 6.50 *10^{-13} J$

Explanation:

To solve this question we are going to be using the concept of conservation of momentum which is mathematically represented as

$m_1v_1 +m_2v_2 +m_3v_3 = 0$

From the question

Mass of unstable atomic nucleus $m_0= 1.58 *10^{-26} kg$

Mass of x-axis particle $m_1 = 8.44* 10^{-27} kg$

Velocity of x-axis particle $v_ 1 = 4.00* 10^6 m/s$

Mass of y-axis particle $m_2 =5.20* 10^{-27} kg$

Velocity of y-axis particle $v_2 = 6.00 *10^6 m/s$

Since the initial mass is known we can obtain the mass of the third particle

Using the conservation of mass mathematical expression

$m_3 = m_0 - (m_1 +m_2)$

$m_3 = 1.58 *10^{-26} - ( 8.44* 10^{-27} + 5.20* 10^-27)$

$= 2.16 *10 ^{-27}kg$

Using the mathematical expression above for conservation of momentum

$(8.44 * 10^{-27})( 4.00* 10^6i) + (5.20 *10^{-27})(6.00* 10^6j) +(2.16*0^{-27})v_3 =0$

$v_3 = (-15.629*10^{6}i -14.45*10^{6}j)\ m/s$

The above is the vector solution now to obtain the value in numeric form we evaluate the magnitude of the vector

$Mag = \sqrt{(15.629*10^6)^2 +(14.45*10^6)^2}$

$= 21.285*10^6 \ m/s$

To obtain the total increase in kinetic energy which is mathematically represented as

$E = \frac{1}{2}m_1v_1 + \frac{1}{2}m_2v_2 + \frac{1}{2} m_3 v_3$

$E =\frac{1}{2} [(5.20*10^{-27})(6.0*0^{6})^2 + (8.44*10^{-27})(4.00*10^6)^2+\\\\(2.16*10^{-27})(21.285*10^6)^2]$

$= 6.50 *10^{-13} J$

5 0

3 years ago