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denpristay [2]
3 years ago
13

Particle q₁ has a charge of 2.7 μC and a velocity of 773 m/s. If it experiences a magnetic force of 5.75 × 10⁻³ N, what is the s

trength of the magnetic field?
A. 0.36 T
B. 1.6 T
C. 2.8 T

In the same magnetic field, particle q₂ has a charge of 42.0 μC and a velocity of 1.21 × 10³ m/s. What is the magnitude of the magnetic force exerted on particle 2?
A. 0.042 N
B. 0.12 N
C. 0.14 N

Physics
2 answers:
Ne4ueva [31]3 years ago
5 0
The intensity of the magnetic force F experienced by a charge q moving with speed v in a magnetic field of intensity B is equal to
F=qvB \sin \theta
where \theta is the angle between the directions of v and B.

1) Re-arranging the previous formula, we can calculate the value of the magnetic field intensity. The charge is q=2.7 \mu C=2.7 \cdot 10^{-6}C. In this case, v and B are perpendicular, so \theta=90^{\circ}, therefore we have:
B= \frac{F}{qv \sin \theta} = \frac{5.75 \cdot 10^{-3}N}{(2.7 \cdot 10^{-6}C)(773m/s)\sin 90^{\circ}}=2.8 T

2) In this second case, the angle between v and B is \theta=55^{\circ}. The charge is now q=42.0 \mu C=42.0 \cdot 10^{-6}C, and the magnetic field is the one we found in the previous part, B=2.8 T, so we can find the intensity of the force experienced by this second charge:
F=qvB \sin \theta=(42\cdot 10^{-6}C)(1.21 \cdot 10^3 m/s)(2.8 T)(\sin 55^{\circ})=0.12 N
STatiana [176]3 years ago
5 0

Answer:

1. 2.8

2. 0.12

Explanation:

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(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)

There are only two forces acting on the balloon, if we neglect air resistance:

- The weight of the balloon, labelled with W, whose magnitude is

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where m is the mass of the balloon+the helium gas inside and g is the acceleration due to gravity, and whose direction is downward

- The Buoyant force, labelled with B, whose magnitude is

B=\rho_a V g

where \rho_a is the air density, V is the volume of the balloon and g the acceleration due to gravity, and where the direction is upward

(b) 4159 N

The buoyant force is given by

B=\rho_a V g

where \rho_a is the air density, V is the volume of the balloon and g the acceleration due to gravity.

In this case we have

\rho_a = 1.29 kg/m^3 is the air density

V=329 m^3 is the volume of the balloon

g = 9.8 m/s^2 is the acceleration due to gravity

So the buoyant force is

B=(1.29 kg/m^3)(329 m^3)(9.8 m/s^2)=4159 N

(c) 1524 N

The mass of the helium gas inside the balloon is

m_h=\rho_h V=(0.179 kg/m^3)(329 m^3)=59 kg

where \rho_h is the helium density; so we the total mass of the balloon+helium gas inside is

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So now we can find the weight of the balloon:

W=mg=(269 kg)(9.8 m/s^2)=2635 N

And so, the net force on the balloon is

F=B-W=4159 N-2635 N=1524 N

(d) The balloon will rise

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(e) 155 kg

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where m' is the additional mass. Re-arranging the equation for m', we find

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B=\rho_a V g

So, at a certain altitude h, the buoyant force will be no longer greater than the weight of the balloon, therefore the net force will become zero and the balloon will no longer rise.

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