Question 1

The figure shows the motion of electrons in a wire which is near the N pole of a magnet. The wire will be pushed:

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Question 2

In a certain mass spectrograph, an ion beam passes through a velocity filter consisting of mutually perpendicular fields  The beam then enters a region of another magnetic field  perpendicular to the beam. The radius of curvature of the resulting ion beam is proportional to:

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Question 3

The diagram shows a straight wire carrying current i in a uniform magnetic field. The magnetic force on the wire is indicated by an arrow but the magnetic field is not shown. Of the following possibilities, the direction of the magnetic field is:

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Question 4

An electron and a proton are both initially moving with the same speed and in the same direction at 90˚ to the same uniform magnetic field. They experience magnetic forces, which are initially:

Accepted Answer

The magnetic force on a charged particle is given by $F = qvB\sin(\theta)$, where $q$ is the charge, $v$ is the velocity, $B$ is the magnetic field strength, and $\theta$ is the angle between the velocity and the magnetic field. Since the electron and proton have equal and opposite charges and are moving at the same speed in the same direction relative to the magnetic field, the magnitudes of their forces will be equal, but the directions will be opposite due to their opposite charges.

Question 5

A loop of wire carrying a current of 2.0 A is in the shape of a right triangle with two equal sides, each 15 cm long. A 0.7 T uniform magnetic field is parallel to the hypotenuse. The resultant magnetic force on the two sides has a magnitude of:

Accepted Answer

The magnetic force on a current-carrying wire is given by F = I * L * B * sin(θ). Since the magnetic field is parallel to the hypotenuse, it is perpendicular to the equal sides of the triangle, making θ = 90° for these sides. However, the forces on the two equal sides are equal in magnitude and opposite in direction, thus canceling each other out, resulting in a net force of 0.

Question 6

Electrons (mass m, charge -e) are accelerated from rest through a potential difference V and are then deflected by a magnetic field  that is perpendicular to their velocity. The radius of the resulting electron trajectory is:

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Question 7

A uniform magnetic field is directed into the page. A charged particle, moving in the plane of the page, follows a clockwise spiral of decreasing radius as shown. A reasonable explanation is:

Accepted Answer

Since the magnetic field is into the page and the charge is moving clockwise, it means that the magnetic force must be directed towards the center of the spiral. Since the radius of the spiral is decreasing, this means that the magnitude of the magnetic force is decreasing. Therefore, the charge must be slowing down. Furthermore, since the force is directed towards the center of the spiral, the charge must be negative (as the force on a positive charge would be directed outwards). Thus, the best choice is B.

Question 8

At one instant an electron is moving in the positive x direction along the x axis in a region where there is a uniform magnetic field in the positive z direction. When viewed from a point on the positive z axis, it subsequent motion is:

Accepted Answer

Since the magnetic field is in the positive z direction, the force on the electron will be perpendicular to both the velocity vector (in the positive x direction) and the magnetic field vector (in the positive z direction), and will point towards the positive y direction. This means that the electron will undergo circular motion in the xy plane, with a counterclockwise direction of rotation (as viewed from above the plane).

Question 9

The figure shows a uniform magnetic field  directed to the left and a wire carrying a current into the page. The magnetic force acting on the wire is:

Accepted Answer

According to the right-hand rule for magnetic force, if the current is into the page and the magnetic field is directed to the left, the resulting force will be perpendicular to both and directed upwards, towards the top of the page.

Question 10

The current is from left to right in the conductor shown. The magnetic field is into the page and point S is at a higher potential than point T. The charge carriers are:

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Question 11

An electron is launched with velocity in a uniform magnetic field The angle $\theta$ between is between 0 and 90^o. As a result, the electron follows a helix, its velocity vector returning to its initial value in a time interval of:

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Question 12

The magnetic torque exerted on a flat current-carrying loop of wire by a uniform magnetic field  is:

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Question 13

A loop of wire carrying a current of 2.0 A is in the shape of a right triangle with two equal sides, each 15 cm long. A 0.7 T uniform magnetic field is in the plane of the triangle and is perpendicular to the hypotenuse. The resultant magnetic force on the two sides has a magnitude of:

Accepted Answer

We can use the formula for the magnetic force on a current-carrying wire:
$F = ILB\sin	heta$
where $I$ is the current, $L$ is the length of the wire, $B$ is the magnetic field, and $	heta$ is the angle between the wire and the magnetic field.

For the two equal sides of the triangle, $	heta = 90^\circ$, so $\sin	heta = 1$. The length of each side is 15 cm, so the total length of the wire is $L = 30$ cm = 0.3 m.

The force on each of the two sides is then:
$F = ILB\sin	heta = (2.0	ext{ A})(0.3	ext{ m})(0.7	ext{ T})(1) = 0.42$ N

Since the two forces are at right angles to each other, the resultant force is the vector sum of the two forces, which has magnitude:
$F_{net} = \sqrt{F^2 + F^2} = \sqrt{2F^2} = \sqrt{2}(0.42	ext{ N}) = 0.60$ N

Therefore, the closest choice is C, 0.30 N.

Question 14

An ion with a charge of +3.25 *10⁻¹⁹ C is in region where a uniform electric field of 5 * 10^4.V/m is perpendicular to a uniform magnetic field of 0.8 T. If its acceleration is zero then its speed must be:

Accepted Answer

The magnetic force on the ion is given by:$$F_m = qvB$$where:* q is the charge of the ion* v is the speed of the ion* B is the magnetic field strengthThe electric force on the ion is given by:$$F_e = qE$$where:* q is the charge of the ion* E is the electric field strengthIf the ion is in equilibrium, then the magnetic force and the electric force must be equal in magnitude and opposite in direction. Therefore:$$F_m = F_e$$$$qvB = qE$$$$v = \frac{E}{B}$$Substituting the given values into this equation, we get:$$v = \frac{5 	imes 10^4 	ext{ V/m}}{0.8 	ext{ T}} = 6.3 	imes 10^5 	ext{ m/s}$$Therefore, the speed of the ion must be 6.3 × 10^5 m/s.

Question 15

The diagram shows a straight wire carrying a flow of electrons into the page. The wire is between the poles of a permanent magnet. The direction of the magnetic force exerted on the wire is:

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Question 16

A cyclotron operates with a given magnetic field and at a given frequency. If R denotes the radius of the final orbit, the final particle energy is proportional to:

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Question 17

A square loop of wire lies in the plane of the page and carries a current I as shown. There is a uniform magnetic field  parallel to the side MK as indicated. The loop will tend to rotate:

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Question 18

An electron and a proton both each travel with equal speeds around circular orbits in the same uniform magnetic field, as shown in the diagram (not to scale). The field is into the page on the diagram. Because the electron is less massive than the proton and because the electron is negatively charged and the proton is positively charged:

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Question 19

J. J. Thomson's experiment, involving the motion of an electron beam in mutually perpendicular  fields, gave the value of:

Accepted Answer

J.J. Thomson's experiment involving the motion of an electron beam in mutually perpendicular electric and magnetic fields gave the value of the charge/mass ratio of the electron. This was the first time the electron's charge to mass ratio was measured, and it played a crucial role in the development of the atomic model.

Question 20

A current is clockwise around the outside edge of this page and a uniform magnetic field is directed parallel to the page, from left to right. If the magnetic force is the only force acting on the page, the page will turn so the right edge:

Accepted Answer

The right-hand rule indicates that the force on the right edge of the page will be directed towards you when the current is clockwise and the magnetic field is from left to right.

Quiz 28: Magnetic Fields

The figure shows the motion of electrons in a wire which is near the N pole of a magnet. The wire will be pushed:

In a certain mass spectrograph, an ion beam passes through a velocity filter consisting of mutually perpendicular fields The beam then enters a region of another magnetic field perpendicular to the beam. The radius of curvature of the resulting ion beam is proportional to:

The diagram shows a straight wire carrying current i in a uniform magnetic field. The magnetic force on the wire is indicated by an arrow but the magnetic field is not shown. Of the following possibilities, the direction of the magnetic field is:

An electron and a proton are both initially moving with the same speed and in the same direction at 90˚ to the same uniform magnetic field. They experience magnetic forces, which are initially:

A loop of wire carrying a current of 2.0 A is in the shape of a right triangle with two equal sides, each 15 cm long. A 0.7 T uniform magnetic field is parallel to the hypotenuse. The resultant magnetic force on the two sides has a magnitude of:

Electrons (mass m, charge -e) are accelerated from rest through a potential difference V and are then deflected by a magnetic field that is perpendicular to their velocity. The radius of the resulting electron trajectory is:

A uniform magnetic field is directed into the page. A charged particle, moving in the plane of the page, follows a clockwise spiral of decreasing radius as shown. A reasonable explanation is:

At one instant an electron is moving in the positive x direction along the x axis in a region where there is a uniform magnetic field in the positive z direction. When viewed from a point on the positive z axis, it subsequent motion is:

The figure shows a uniform magnetic field directed to the left and a wire carrying a current into the page. The magnetic force acting on the wire is:

The current is from left to right in the conductor shown. The magnetic field is into the page and point S is at a higher potential than point T. The charge carriers are:

The magnetic torque exerted on a flat current-carrying loop of wire by a uniform magnetic field is:

A loop of wire carrying a current of 2.0 A is in the shape of a right triangle with two equal sides, each 15 cm long. A 0.7 T uniform magnetic field is in the plane of the triangle and is perpendicular to the hypotenuse. The resultant magnetic force on the two sides has a magnitude of:

An ion with a charge of +3.25 10⁻¹⁹ C is in region where a uniform electric field of 5 10^4.V/m is perpendicular to a uniform magnetic field of 0.8 T. If its acceleration is zero then its speed must be:

A cyclotron operates with a given magnetic field and at a given frequency. If R denotes the radius of the final orbit, the final particle energy is proportional to:

A square loop of wire lies in the plane of the page and carries a current I as shown. There is a uniform magnetic field parallel to the side MK as indicated. The loop will tend to rotate:

J. J. Thomson's experiment, involving the motion of an electron beam in mutually perpendicular fields, gave the value of:

A current is clockwise around the outside edge of this page and a uniform magnetic field is directed parallel to the page, from left to right. If the magnetic force is the only force acting on the page, the page will turn so the right edge:

Quiz 28: Magnetic Fields

The figure shows the motion of electrons in a wire which is near the N pole of a magnet. The wire will be pushed:

In a certain mass spectrograph, an ion beam passes through a velocity filter consisting of mutually perpendicular fields The beam then enters a region of another magnetic field perpendicular to the beam. The radius of curvature of the resulting ion beam is proportional to:

The diagram shows a straight wire carrying current i in a uniform magnetic field. The magnetic force on the wire is indicated by an arrow but the magnetic field is not shown. Of the following possibilities, the direction of the magnetic field is:

An electron and a proton are both initially moving with the same speed and in the same direction at 90˚ to the same uniform magnetic field. They experience magnetic forces, which are initially:

A loop of wire carrying a current of 2.0 A is in the shape of a right triangle with two equal sides, each 15 cm long. A 0.7 T uniform magnetic field is parallel to the hypotenuse. The resultant magnetic force on the two sides has a magnitude of:

Electrons (mass m, charge -e) are accelerated from rest through a potential difference V and are then deflected by a magnetic field that is perpendicular to their velocity. The radius of the resulting electron trajectory is:

A uniform magnetic field is directed into the page. A charged particle, moving in the plane of the page, follows a clockwise spiral of decreasing radius as shown. A reasonable explanation is:

At one instant an electron is moving in the positive x direction along the x axis in a region where there is a uniform magnetic field in the positive z direction. When viewed from a point on the positive z axis, it subsequent motion is:

The figure shows a uniform magnetic field directed to the left and a wire carrying a current into the page. The magnetic force acting on the wire is:

The current is from left to right in the conductor shown. The magnetic field is into the page and point S is at a higher potential than point T. The charge carriers are:

An electron is launched with velocity in a uniform magnetic field The angle θ\thetaθ between is between 0 and 90o. As a result, the electron follows a helix, its velocity vector returning to its initial value in a time interval of:

The magnetic torque exerted on a flat current-carrying loop of wire by a uniform magnetic field is:

A loop of wire carrying a current of 2.0 A is in the shape of a right triangle with two equal sides, each 15 cm long. A 0.7 T uniform magnetic field is in the plane of the triangle and is perpendicular to the hypotenuse. The resultant magnetic force on the two sides has a magnitude of:

An ion with a charge of +3.25 *10−19 C is in region where a uniform electric field of 5 * 104. V/m is perpendicular to a uniform magnetic field of 0.8 T. If its acceleration is zero then its speed must be:

The diagram shows a straight wire carrying a flow of electrons into the page. The wire is between the poles of a permanent magnet. The direction of the magnetic force exerted on the wire is:

A cyclotron operates with a given magnetic field and at a given frequency. If R denotes the radius of the final orbit, the final particle energy is proportional to:

A square loop of wire lies in the plane of the page and carries a current I as shown. There is a uniform magnetic field parallel to the side MK as indicated. The loop will tend to rotate:

J. J. Thomson's experiment, involving the motion of an electron beam in mutually perpendicular fields, gave the value of:

A current is clockwise around the outside edge of this page and a uniform magnetic field is directed parallel to the page, from left to right. If the magnetic force is the only force acting on the page, the page will turn so the right edge:

An ion with a charge of +3.25 10⁻¹⁹ C is in region where a uniform electric field of 5 10^4.V/m is perpendicular to a uniform magnetic field of 0.8 T. If its acceleration is zero then its speed must be: