High School

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(1) Cosmic rays are charged particles like protons and electrons that travel through the universe at very high speeds. A cosmic ray travels through Earth's magnetic field, [tex]B_E[/tex], which points down. Describe two directions the cosmic ray could be traveling if it experiences zero magnetic force due to [tex]B_E[/tex].

(2) Explain why the cosmic ray feels no magnetic force when moving in either of those two directions.

(3) Consider a scenario where the cosmic ray is an electron moving east. Describe a situation that would result in a magnetic force. Write a one-sentence description, for example: "The electron is traveling east in a magnetic field [tex]B[/tex] that points down, and the field exerts a force [tex]F[/tex] directed north." Then, sketch a diagram to illustrate your answer with vectors labeled [tex]B_E[/tex], [tex]v[/tex], and [tex]F[/tex], and axes showing four of the six possible directions. Use standard symbols for into the page (×) and out of the page (·).

(4) The speed of your cosmic ray is 4.56% the speed of light, and [tex]B_E = 45.3 \, \mu T[/tex]. Calculate the magnetic force Earth's magnetic field exerts on the cosmic ray.

For the rest of the assignment, explore the magnetic fields created by current-carrying wires in three different configurations of increasing field strength.

(5) A long, straight wire carries a current of 6.78 A. Calculate how close you'd have to get to this wire for its field [tex]B_L[/tex] to rise to the magnitude [tex]B_E[/tex] from question 4. Check: does this distance give the right field?

(6) The equations for calculating the magnetic fields of coils of wire or solenoids contain the symbols [tex]N[/tex] and [tex]n[/tex]. Describe the physical difference between the quantities represented by [tex]N[/tex] and [tex]n[/tex].

(7) You can concentrate the magnetic field of a long wire by coiling it up. A circular coil has 18 turns and a radius of 16.5 cm. Calculate the magnetic field strength [tex]B_C[/tex] in mT at the center of this coil when it carries the current chosen in question 5.

(8) A solenoid can create a stronger magnetic field than a circular coil for the same amount of current. A solenoid is constructed by wrapping wire with a diameter of 0.342 mm around a hollow cylinder longer than a soda can but shorter than your arm. With a wire length of 8.38 inches, calculate the maximum whole number of turns of wire you could wrap around the cylinder in one layer.

(9) Calculate the magnetic field [tex]B_S[/tex] inside the solenoid in mT, assuming it has the current chosen in question 5. Check: how should this compare to [tex]B_L[/tex] and [tex]B_C[/tex], since the current is the same?

(10) Calculate the current [tex]I_L[/tex] the long, straight wire would have to carry to match the solenoid's magnetic field [tex]B_S[/tex] at the distance found in question 5. Can you think of a way to verify this?

Answer :

(1) Two directions in which a cosmic ray could be traveling to experience zero magnetic force due to Earth's magnetic field (BE) are: North-South direction and Vertical direction

a) North-South direction: If the cosmic ray is traveling directly along the magnetic field lines, i.e., in a direction parallel to the field lines, it will experience zero magnetic force.

b) Vertical direction: If the cosmic ray is traveling vertically, perpendicular to the Earth's magnetic field lines, it will also experience zero magnetic force.

(2) The cosmic ray feels no magnetic force when moving in either of those two directions because the magnetic force on a charged particle depends on the cross product of its velocity vector and the magnetic field vector. When the velocity vector is parallel or anti-parallel to the magnetic field vector, the cross product becomes zero, resulting in no magnetic force acting on the particle.

(3) The electron is traveling eastward in a magnetic field B that points downward, and the field exerts a force F directed southward.

(4) Given:

Speed of the cosmic ray (v) = 0.0456 * c (speed of light) = 0.0456 * 3 × 10^8 m/s = 1.368 × 10^7 m/s

Magnetic field strength (BE) = 45.3 µT = 45.3 × 10^(-6) T

The magnetic force (F) on a charged particle moving through a magnetic field is given by the equation: F = q * v * B, where q is the charge of the particle.

As the cosmic ray is an electron with charge q = -1.6 × 10^(-19) C, we can calculate the magnetic force:

F = (-1.6 × 10^(-19) C) * (1.368 × 10^7 m/s) * (45.3 × 10^(-6) T)

F ≈ -1.107 × 10^(-15) N (southward direction, as the charge is negative)

Therefore, the Earth's magnetic field exerts a magnetic force of approximately -1.107 × 10^(-15) N on the cosmic ray.

(5) To calculate the distance at which the magnetic field (BL) of the long, straight wire carrying a current of 6.78 A matches the magnitude of Earth's magnetic field (BE):

Using the formula for the magnetic field created by a long, straight wire at a distance r from the wire:

BL = (μ₀ * I) / (2 * π * r), where μ₀ is the permeability of free space, I is the current, and r is the distance from the wire.

Setting BL equal to BE:

(μ₀ * I) / (2 * π * r) = BE

Solving for r:

r = (μ₀ * I) / (2 * π * BE)

Given:

μ₀ = 4π × 10^(-7) T·m/A (permeability of free space)

I = 6.78 A

BE = 45.3 µT = 45.3 × 10^(-6) T

Substituting the values:

r = (4π × 10^(-7) T·m/A * 6.78 A) / (2 * π * 45.3 × 10^(-6) T)

r ≈ 0.299 m

Therefore, you would need to get approximately 0.

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