Question 1

A current in a long,straight wire produces a magnetic field.The magnetic field lines:

Accepted Answer

The magnetic field produced by a current in a straight wire forms circles that go around the wire, according to the right-hand rule for magnetic fields.

Question 2

In a mass spectrometer,an ion will have a smaller radius for its circular path if:

Accepted Answer

The answer of In a mass spectrometer,an ion will have...

Question 3

Two long parallel wires 20 cm apart carry currents of 5.0 A and 8.0 A in the same direction.Is there any point between the two wires where the magnetic field is zero?

Accepted Answer

The magnetic field due to each wire at a point between them is given by:
B = (μ₀/4π) * I / d
where I is the current, d is the distance from the wire, and μ₀ is the permeability of free space.
At the midpoint between the wires, the fields due to the two wires are in opposite directions and have equal magnitudes, resulting in a net magnetic field of zero.
At a distance of 12 cm from the 5-A wire, the fields due to the two wires are not equal and opposite, so the net magnetic field is not zero.
However, at a distance of 7.7 cm from the 5-A wire, the fields due to the two wires are again equal and opposite, resulting in a net magnetic field of zero. Therefore, the correct answer is (C).

Question 4

A proton with initial kinetic energy E is moving in circular motion in a uniform magnetic field.When it has completed one eighth of a revolution,what is its kinetic energy?

Accepted Answer

The answer of A proton with initial kinetic energy E...

Question 5

An incredible amount of electrical energy passes down the funnel of a large tornado every second.Measurements taken in Oklahoma at a distance of 9.00 km from a large tornado showed an almost constant magnetic field of 1.50 x 10^-8 T associated with the tornado.What was the average current going down the funnel? (µ₀ = 4p x 10^-7 T*m/A)

Accepted Answer

The magnetic field (B) around a long straight conductor is given by B = (µ₀ * I) / (2π * r). Rearranging for I (current), we get I = (B * 2π * r) / µ₀. Substituting the given values: B = 1.50 x 10⁻⁸ T, r = 9.00 km = 9000 m, and µ₀ = 4π x 10⁻⁷ Tm/A, we find I = (1.50 x 10⁻⁸ T * 2π * 9000 m) / (4π x 10⁻⁷ Tm/A) = 675 A.

Question 6

A proton,mass 1.67 x 10^-27kg and charge +1.6 x 10^-19 C,moves in a circular orbit perpendicular to a uniform magnetic field of 0.75 T.Find the time for the proton to make one complete circular orbit.

Accepted Answer

The time for one complete circular orbit, also known as the period, can be found using the formula:
T = 2πm/qB
where T is the period, m is the mass of the particle, q is its charge, and B is the magnetic field.
Substituting the given values, we get:
T = 2π(1.67 x 10^-27 kg)/(1.6 x 10^-19C)(0.75 T)
T = 8.7 x 10^-8 s
Therefore, the correct choice is B).

Question 7

An electron is moving at a speed of 6.0 x 10⁶ m/s at an angle of 30 $\degree$ with respect to a uniform magnetic field of 8.0 x 10^-4 T.What is the radius of the resulting helical path? (m_e = 9.11 x 10^-31 kg,q_e = 1.6 x 10^-19 C)

Accepted Answer

The answer of An electron is moving at a speed...

Question 8

Two long parallel wires 40 cm apart are carrying currents of 10 A and 20 A in the same direction.What is the magnitude of the magnetic field halfway between the wires?

Accepted Answer

The magnetic field due to a long straight wire carrying current is given by:B = (μ0 * I) / (2πr)where:μ0 is the permeability of free space (4π x 10^-7 T m/A)I is the current in the wire (in amperes)r is the distance from the wire (in meters)In this problem, we have two wires carrying currents of 10 A and 20 A, respectively. The wires are 40 cm apart, and we want to find the magnetic field halfway between them. The distance from each wire to the midpoint is 20 cm = 0.2 m.The magnetic field due to the first wire is:B1 = (μ0 * I1) / (2πr1)= (4π x 10^-7 T m/A) * (10 A) / (2π * 0.2 m)= 1.0 x 10^-5 TThe magnetic field due to the second wire is:B2 = (μ0 * I2) / (2πr2)= (4π x 10^-7 T m/A) * (20 A) / (2π * 0.2 m)= 2.0 x 10^-5 TThe total magnetic field at the midpoint is the sum of the magnetic fields due to each wire:B = B1 + B2= 1.0 x 10^-5 T + 2.0 x 10^-5 T= 3.0 x 10^-5 TTherefore, the magnitude of the magnetic field halfway between the wires is 3.0 x 10^-5 T.

Question 9

The path of a charged particle moving parallel to a uniform magnetic field will be a:

Accepted Answer

When a charged particle moves parallel to a uniform magnetic field, it experiences no magnetic force, thus it continues to move in a straight line.

Question 10

Magnetism had been a known phenomenon for some time before its relation to electric currents was found.That a current in a wire produces a magnetic field was discovered by:

Accepted Answer

The discovery that a current in a wire produces a magnetic field was made by Hans Christian Oersted in 1820. Maxwell, Ampere, and Tesla made significant contributions to the understanding of electromagnetism, but they did not discover the relation between electric currents and magnetism.

Question 11

A proton,which moves perpendicular to a magnetic field of 1.2 T in a circular path of radius 0.080 m,has what speed? (q_p = 1.6 x 10^-19 C and m_p = 1.67 x 10^-27 kg)

Accepted Answer

The answer of A proton,which moves perpendicular to a magnetic...

Question 12

A proton moving with a speed of 3.0 x 10⁵ m/s perpendicular to a uniform magnetic field of 0.20 T will follow which of the paths described below? (q_p = 1.6 x 10^-19 C and m_p = 1.67 x 10^-27 kg)

Accepted Answer

The answer of A proton moving with a speed of...

Question 13

Two singly ionized isotopes,X and Y,of the same element move with the same speed perpendicular to a uniform magnetic field.Isotope X follows a path of radius 3.35 cm while isotope Y moves along a path 3.43 cm in radius.What is the ratio of the two isotope masses,m_X/m_Y?

Accepted Answer

The radius of the circular path is inversely proportional to the mass of the ion. Therefore, mX/mY = rY/rX = 3.43 cm / 3.35 cm = 1.02

Question 14

Niobium metal becomes a superconductor (with electrical resistance equal to zero)when cooled below 9 K.If superconductivity is destroyed when the surface magnetic field exceeds 0.100 T,determine the maximum current a 4.00-mm-diameter niobium wire can carry and remain superconducting.(µ₀ = 4p x 10^-7 T*m/A)

Accepted Answer

The answer of Niobium metal becomes a superconductor (with electrical...

Question 15

When a magnetic field causes a charged particle to move in a circular path,the only quantity listed below which the magnetic force changes significantly as the particle goes around in a circle is the particle's:

Accepted Answer

The magnetic force exerted on a charged particle moving in a magnetic field is given by F = q(v x B), where v is the velocity of the particle and B is the magnetic field. This force is perpendicular to both the velocity and the magnetic field, and thus it causes the charged particle to move in a circular path. 
The radius of the circular path depends on the velocity, mass, and charge of the particle, as well as the strength of the magnetic field. However, the momentum of the particle (p=mv) is conserved in the absence of external forces, and thus it remains constant as the particle moves in a circle. Therefore, the magnetic force does not change the momentum of the particle significantly as it goes around in a circle.

Question 16

A deuteron,with the same charge but twice the mass of a proton,moves with a speed of 3.0 ´ 10⁵m/s perpendicular to a uniform magnetic field of 0.20 T.Which of the paths described below would it follow? (q_p = 1.6 x 10^-19 C and m_d = 3.34 x 10^-27 kg)

Accepted Answer

The formula for the radius of the circular path of a charged particle in a magnetic field is:
r = mv / (qB)
where r is the radius of the circular path, m is the mass of the particle, v is its velocity, q is its charge, and B is the magnetic field strength.
Substituting the values given:
r = (2 * m * v) / (q * B)
r = (2 * 3.34 x 10^-27 kg * 3.0 x 10^5 m/s) / (1.6 x 10^-19 C * 0.20 T)
r = 3.1 cm
Therefore, the deuteron would follow a circular path with a radius of 3.1 cm. Choice C is the correct answer.

Question 17

A 100-m-long wire carrying a current of 4.0 A will be accompanied by a magnetic field of what strength at a distance of 0.050 m from the wire? (magnetic permeability in empty space µ₀= 4p x 10^-7 T*m/A)

Accepted Answer

The magnetic field strength (B) at a distance (r) from a long straight wire carrying a current (I) can be calculated using the formula:

B = (µ0 × I) / (2π × r)

where µ0 is the magnetic permeability in empty space, which is given as 4π × 10^-7 T·m/A in the question.

Plugging in the given values, we get:

B = (4π × 10^-7 T·m/A × 4.0 A) / (2π × 0.050 m)
B = 1.6 × 10^-5 T

Therefore, the correct choice is C, 1.6 × 10^-5 T.

Question 18

The current in a long wire creates a magnetic field in the region around the wire.How is the strength of the field at distance r from the wire center related to the magnitude of the field?

Accepted Answer

The strength of the magnetic field created by a current-carrying wire is inversely proportional to the distance from the wire center. This is known as the inverse-square law, which states that the intensity of a physical quantity decreases with the square of the distance from the source. Therefore, as the distance from the wire increases, the magnetic field strength will decrease.

Question 19

If a charged particle is moving in a uniform magnetic field,its path can be:

Accepted Answer

The path of a charged particle moving in a uniform magnetic field can be a straight line (if it moves parallel to the field lines), a circle (if it moves perpendicular to the field lines), or a helix (if it moves at an angle to the field lines).

Question 20

At the Fermilab accelerator in Weston,Illinois,singly-charged ions with momentum 4.8 x 10^-16 kg*m/s are held in a circular orbit of radius 1 km by an upward magnetic field.What B-field must be used to maintain the ions in this orbit? (q_ion= 1.6 x 10^-19 C)

Accepted Answer

The answer of At the Fermilab accelerator in Weston,Illinois,singly-charged ions...

Quiz 19: Magnetism

A current in a long,straight wire produces a magnetic field.The magnetic field lines:

In a mass spectrometer,an ion will have a smaller radius for its circular path if:

Two long parallel wires 20 cm apart carry currents of 5.0 A and 8.0 A in the same direction.Is there any point between the two wires where the magnetic field is zero?

A proton with initial kinetic energy E is moving in circular motion in a uniform magnetic field.When it has completed one eighth of a revolution,what is its kinetic energy?

A proton,mass 1.67 x 10-27 kg and charge +1.6 x 10-19 C,moves in a circular orbit perpendicular to a uniform magnetic field of 0.75 T.Find the time for the proton to make one complete circular orbit.

An electron is moving at a speed of 6.0 x 106 m/s at an angle of 30 °\degree° with respect to a uniform magnetic field of 8.0 x 10-4 T.What is the radius of the resulting helical path? (me = 9.11 x 10-31 kg,qe = 1.6 x 10-19 C)