What does the model show?

A vertical straight conductor X of length 0.5m is held along the positive X-axis and situated in a uniform horizontal magnetic f

k0ka [10]

Answer:

i. 0.2 N ii. 30°

Explanation:

(i) Calculate the magnitude and direction of force on X, when a current of 4A is passed through it.

The magnetic force F = BILsinФ where B = magnetic field strength = 0.1 T, I = current = 4 A and L= length of conductor = 0.5 m. Since the conductor X of length 0.5m is held along the positive X-axis and situated in a uniform horizontal magnetic field of 0.1T which is pointing towards the positive Y-axis, both B and L are perpendicular to each other. So, Ф = 90°

So, F = BILsinФ

F = 0.1 T × 4 A ×0.5 m × sin90°

F = 0.1 T × 4 A ×0.5 m × 1

F = 0.2 N

(ii) Through what angle must X be turned in a vertical plane so that the force on X is halved

If F' = BILsinФ' where Ф'=the new angle, and BIL = F

F'/F = sinФ'

Since F'/F = 1/2

sinФ' = 1/2

Ф' = sin⁻¹(1/2) = 30°

4 0

3 years ago

A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences

NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, $q=3\times 10^{-6}\ C$

Mass of charge particle, $m=2\times 10^{-6}\ C$

Speed, $v=5\times 10^{6}\ m/s$

Acceleration, $a=3\times 10^{4}\ m/s^2$

We need to find the minimum magnetic field that would produce such an acceleration. So,

$ma=qvB\ sin\theta$

For minimum magnetic field,

$ma=qvB$

$B=\dfrac{ma}{qv}$

$B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}$

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0

3 years ago

A metal smith pours 3.00 kg of lead shot at 99oc into 1.00 kg of water at 25oc in an isolated container. what is the final tempe

cluponka [151]

The answer is in attachment.

7 0

3 years ago

Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure.

Vesnalui [34]

The angle of the planet is mathematically given as

dY= 704 degrees

<h3>What angle has planet Y rotated through during this time?</h3>

With Kepler's third rule, which states that a planet's orbit squared is a function of cubed radius, we can prove that this is the case.

Generally, the equation for the period is mathematically given as

(periodX / periodY)^2 = (radius X / radius Y)^3

Therefore

(pX / pY)^2 = 4^3

(pX / pY)^2 = 64

\sqrt{(pX / pY )^2}= \sqrt{64}

(pX / pY=8

In conclusion, Because it takes 8 times longer to complete one orbit on planet X, planet Y travels 8 times farther than planet X does in the same time period...

planet Y travels ;

dY=8 * 88.0

dY= 704 degrees