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DiKsa [7]
4 years ago
14

Particle accelerators, such as the Large Hadron Collider, use magnetic fields to steer charged particles around a ring Consider

a proton ring with 36 identical bending magnets connected by straight segments. The protons move along a 1.0-m-long circular arc as they pass through each magnet. Part A What magnetic field strength is needed in each magnet to steer protons around the ring with a speed of 2.0 x 10 m/s? Assume that the field is uniform inside the magnet, zero outside. Express your answer with the appropriate units
Physics
1 answer:
coldgirl [10]4 years ago
6 0

Answer:

\beta= 3.49x10^{-8}T

Explanation:

The magnetic field can be find using the equation

m*v^2/r=q*v*\beta

You can cancel a element of v'

m*v/r=q*\beta

C=36*1m=2\pi*r

r=\frac{36}{2\pi } =5.7295m

Solve to magnetic field

\beta=\frac{m*v^2}{r*q}

The charge and mass of the proton are:

m_p=1.6x10^{27}kg, q_p=1.6x10^{-19}C

Replacing numeric

\beta=\frac{1.6x10^{-27}kg*2x10m/s}{1.6x10^{-19}C*5.73m}

\beta= 3.49x10^{-8}T

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5 0
4 years ago
Please tell me some streamlined objects
Sauron [17]
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3 0
3 years ago
A mustang, has an average velocity of 33 m/s while covering a course that is 21 miles long. (1 mile = 1609 m). How long did it t
saw5 [17]

Answer:

t = 1023.9 seconds

Explanation:

Given that,

The average velocity of a Mustang, v = 33m/s

Distance, d = 21 miles = 33789 m

Let the driver takes t seconds. So,

Speed = distance/time

t=\dfrac{d}{v}\\\\t=\dfrac{33789}{33}\\\\t=1023.9\ s

So, it will take 1023.9 seconds to complete the course.

3 0
3 years ago
The insulating solid sphere in the previous Example 24.5 has the same total charge Q and radius a as the thin shell in the examp
BaLLatris [955]

Answer and Explanation:

Using Gauss's law,

If r>a

then charge enclosed in all the three cases is same as Q.

So Electric field for all three is same.

So {1,1,1}.

(b) r<a,

Charge enclosed in case of shell is zero since all charge is present on the surface. So E = 0.

Charge enclosed by incase of point charge is Q.

Charge enclosed in case of sphere is Qr3/a3 which is less than Q.

So ranking {2,3,1}

6 0
4 years ago
A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radiu
anyanavicka [17]

Answer:

1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

Explanation:

According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.

As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :

Q₁ = ∫ ρ dV

Here dV is the volume element of sphere of radius r.

Q₁ = ρ x 4π x ∫ r² dr

The limit of integration is from 0 to r as r is less than R.

Q₁ = (4π x ρ x r³ )/3

But volume charge density, ρ = \frac{3Q}{4\pi R^{3} }

So, Q_{1} = \frac{Qr^{3} }{R^{3} }

Applying Gauss law of electrostatics ;

∫ E ds = Q₁/ε₀

Here E is electric field inside the sphere and ds is surface element of sphere of radius r.

Substitute the value of Q₁ in the above equation. Hence,

E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

7 0
4 years ago
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