Question 1

A positively charged particle has a velocity in the negative z direction at point P.The magnetic force on the particle at this point is in the negative y direction.Which one of the following statements about the magnetic field at point P can be determined from this data?

Accepted Answer

A

Question 2

A straight wire is bent into the shape shown.Determine the net magnetic force on the wire when the current I travels in the direction shown in the magnetic field

Accepted Answer

B

Question 3

A charged particle (mass = 4.0 $\mu$g, charge = 5.0 $\mu$C) moves in a region where the only force on it is magnetic.What is the magnitude of the acceleration of the particle at a point where the speed of the particle is 5.0 km/s, the magnitude of the magnetic field is 8.0 mT, and the angle between the direction of the magnetic field and the velocity of the particle is 60$\degree$?

Accepted Answer

B

Question 4

An electron has a velocity of 6.0 *106 m/s in the positive x direction at a point where the magnetic field has the components Bx = 3.0 T, By = 1.5 T, and Bz = 2.0 T.What is the magnitude of the acceleration of the electron at this point?

Accepted Answer

The magnitude of the acceleration of an electron moving in a magnetic field is given by the equation: 
a = (e/m)*(v x B), 
where e is the charge of the electron, m is its mass, v is its velocity, and B is the magnetic field vector. 
Taking the cross product of v and B, we get: 
v x B = (v_y*B_z - v_z*B_y, v_z*B_x - v_x*B_z, v_x*B_y - v_y*B_x)
Plugging in the given values, we get: 
v x B = (-1.5x106*2.0, -2.0x3.0x106, 6.0x106*1.5) = (-3.0, -6.0, 9.0) x 1012 m/s2 
Multiplying by the charge-to-mass ratio of the electron (1.76x1011 C/kg), we get: 
a = (1.76x1011)*(3.0, 6.0, -9.0) x 1012 m/s2 = (5.28, 10.56, -15.84) x 1018 m/s2 
Taking the magnitude of the acceleration, we get: 
|a| = sqrt(5.28^2 + 10.56^2 + 15.84^2) x 1018 m/s2 ≈ 2.6 x 1018 m/s2, 
which is closest to answer choice C.

Question 5

A 2.0-m wire carries a current of 15 A directed along the positive x axis in a region where the magnetic field is uniform and given by B = (30 -40) mT.What is the resulting magnetic force on the wire?

Accepted Answer

The answer of A 2.0-m wire carries a current of...

Question 6

An electron moving in the positive x direction experiences a magnetic force in the positive z direction.If Bx = 0, what is the direction of the magnetic field?

Accepted Answer

The magnetic force on a moving electron is given by the Lorentz force equation: F = q(v x B), where q is the charge of the electron, v is its velocity, and B is the magnetic field. In this case, the electron is moving in the positive x direction and experiences a magnetic force in the positive z direction. This means that the magnetic field must be in the negative y direction, since the cross product of the velocity and magnetic field vectors must be in the positive z direction.

Question 7

What current must be maintained in a square loop (50 cm on a side) to create a torque of 1.0 N . m about an axis through its centre and parallel to one of its sides when a magnetic field of magnitude 70 mT is directed at 40 $\degree$ to the plane of the loop?

Accepted Answer

The answer of What current must be maintained in a...

Question 8

A circular loop (radius = 0.50 m) carries a current of 3.0 A and has unit normal vector of ()/3.What is the x component of the torque on this loop when it is placed in a uniform magnetic field of ()T?

Accepted Answer

The torque on a current loop in a magnetic field is given by the formula τ = IABsinθ, where I is the current, A is the area of the loop, B is the magnetic field strength, and θ is the angle between the magnetic field and the normal to the loop. Since the loop is circular, its area is given by A = πr^2 = π(0.5 m)^2 = 0.785 m^2. The angle between the magnetic field and the normal to the loop is 60 degrees, since the unit normal vector is ()/3 and the magnetic field is uniform. Therefore, sinθ = sin(60°) = √3/2. Substituting the given values, we get τ = (3.0 A)(0.785 m^2)(0.5 T)(√3/2) = 9.4 N m. The x-component of the torque is given by τx = τsinθ = (9.4 N m)(√3/2) ≈ 8.14 N m, which is closest to choice D.

Question 9

A 2.0-C charge moves with a velocity of () m/s and experiences a magnetic force of () N.The x component of the magnetic field is equal to zero.Determine the y component of the magnetic field.

Accepted Answer

We can use the formula F = qvBsinθ where F is the magnetic force, q is the charge, v is the velocity, B is the magnetic field, and θ is the angle between v and B. Since the x component of the magnetic field is zero, we can assume that B is entirely in the y direction. Therefore, θ = 90° and sinθ = 1. Solving for B, we have B = F/(qv). Substituting the given values, we get B = (2.0 C)( )( )/( ) = +3.0 T. Therefore, the answer is choice B.

Question 10

A straight wire carries a current of 40 A in a uniform magnetic field (magnitude = 80 mT).If the force per unit length on this wire is 2.0 N/m, determine the angle between the wire and the magnetic field.

Accepted Answer

The answer of A straight wire carries a current of...

Question 11

A particle (q =-4.0 $\mu$C, m = 5.0 mg) moves in a uniform magnetic field with a velocity having a magnitude of 2.0 km/s and a direction that is 50 $\degree$ away from that of the magnetic field.The particle is observed to have an acceleration with a magnitude of 5.8 m/s2.What is the magnitude of the magnetic field?

Accepted Answer

The answer of A particle (q =-4.0 $\mu$C, m =...

Question 12

A particle (mass = 6.0 mg) moves with a speed of 4.0 km/s in a direction that makes an angle of 37 $\degree$ above the positive x axis in the xy plane.At the instant it enters a magnetic field of (5.0) mT it experiences an acceleration of (8.0) m/s2.What is the charge of the particle?

Accepted Answer

The answer of A particle (mass = 6.0 mg) moves...

Question 13

What is the magnitude of the magnetic force on a charged particle (Q = 5.0 $\mu$C) moving with a speed of 80 km/s in the positive x direction at a point where Bx = 5.0 T, By = -4.0 T, and Bz = 3.0 T?

Accepted Answer

The equation for the magnetic force on a charged particle is:
F = QVB
where Q is the charge of the particle, V is its velocity, and B is the magnetic field.

First, we need to convert the particle's speed from km/s to m/s:
80 km/s = 80,000 m/s

Next, we can plug in the given values for Q, V, and B:
F = (5.0 F1F1F1S1F1F1F10C)(80,000 m/s)(5.0 T)   [for the x-component]
F = (5.0 F1F1F1S1F1F1F10C)(80,000 m/s)(F1F1F1S1F1F1F104.0 T)   [for the y-component]
F = (5.0 F1F1F1S1F1F1F10C)(80,000 m/s)(3.0 T)   [for the z-component]

Simplifying each of these equations, we get:
Fx = 2.0 N
Fy = -4.0 N
Fz = 1.2 N

The magnitude of the force is given by:
|F| = sqrt(Fx^2 + Fy^2 + Fz^2)

Plugging in the values we just calculated, we get:
|F| = sqrt((2.0 N)^2 + (-4.0 N)^2 + (1.2 N)^2)
|F| = sqrt(4.0 N^2 + 16.0 N^2 + 1.44 N^2)
|F| = sqrt(21.44 N^2)
|F| = 4.62 N

Rounding to the nearest tenth, we get:
|F| = 4.6 N

Therefore, the answer is D) 2.0 N.

Question 14

A particle (mass = 2.0 mg, charge = -6.0 $\mu$C) moves in the positive direction along the x axis with a velocity of 3.0 km/s.It enters a magnetic field of () mT.What is the acceleration of the particle?

Accepted Answer

The magnetic force on a charged particle in a magnetic field is given by F = qvBsinθ, where q is the charge of the particle, v is the velocity of the particle, B is the magnetic field, and θ is the angle between the velocity and the magnetic field. In this case, the particle moves in the positive x direction, and the magnetic field is in the negative z direction. Therefore, θ = 90°, and sinθ = 1. Plugging in the values, we get F = (106.0 μC)(3.0 km/s)(0.5 mT)(1) = 1.59 × 10-10 N. The acceleration of the particle is given by a = F/m, where m is the mass of the particle. Plugging in the values, we get a = (1.59 × 10-10 N)/(2.0 × 10-9 kg) = 79.5 m/s2. Converting this to scientific notation and rounding to two significant figures, we get a = 7.9 × 101 m/s2 = 36 m/s2. Therefore, the best choice is A.

Question 15

A segment of wire carries a current of 25 A along the x axis from x = -2.0 m to x = 0 and then along the y axis from y = 0 to y = 3.0 m.In this region of space, the magnetic field is equal to 40 mT in the positive z direction.What is the magnitude of the force on this segment of wire?

Accepted Answer

To find the force on the wire segment, we can use the formula:

F = I L B sinθ

where I is the current, L is the length of the wire segment, B is the magnetic field, and θ is the angle between the wire segment and the magnetic field (which is 90 degrees in this case).

First, let's find the length of the wire segment. Along the x axis, the wire segment extends from x = 102.0 m to x = 0, so its length is 102.0 m. Along the y axis, the wire segment extends from y = 0 to y = 3.0 m, so its length is 3.0 m. The total length of the wire segment is therefore:

L = 102.0 m + 3.0 m = 105.0 m

Next, we can plug in the values for I, L, B, and θ:

F = (25 A)(105.0 m)(40 mT) sin(90°)
  = (25 A)(105.0 m)(0.04 T)
  = 105 N

Therefore, the magnitude of the force on the wire segment is 105 N, which is closest to answer choice D (3.6 N).

Question 16

A current of 4.0 A is maintained in a single circular loop having a circumference of 80 cm.An external magnetic field of 2.0 T is directed so that the angle between the field and the plane of the loop is 20 $\degree$.Determine the magnitude of the torque exerted on the loop by the magnetic forces acting upon it.

Accepted Answer

The torque on a current loop is given by the formula:

τ = NIABsinθ

where N is the number of turns in the loop, I is the current, A is the area of the loop, B is the magnetic field, and θ is the angle between the magnetic field and the normal to the plane of the loop.

In this problem, N = 1 (single loop), I = 4.0 A, A = (πd²)/4 = (π0.8²)/4 = 0.5027 m² (where d is the diameter of the loop), B = 2.0 T, and θ = 20°.

Plugging in the values, we get:

τ = (1)(4.0 A)(0.5027 m²)(2.0 T)sin(20°) 
= 0.38 N·m

So the answer is (C) 0.38 N·m.

Question 17

A wire (mass = 50 g, length = 40 cm) is suspended horizontally by two vertical wires which conduct a current I = 8.0 A, as shown in the figure.The magnetic field in the region is into the paper and has a magnitude of 60 mT.What is the tension in either wire?

Accepted Answer

The answer of A wire (mass = 50 g, length...

Question 18

A uniform magnetic field of 0.50 T is directed along the positive x axis.A proton moving with a speed of 60 km/s enters this field.The helical path followed by the proton shown has a pitch of 5.0 mm.Determine the angle between the magnetic field and the velocity of the proton.

Accepted Answer

The answer of A uniform magnetic field of 0.50 T...

Question 19

A circular coil (radius = 0.40 m) has 160 turns and is in a uniform magnetic field.When the orientation of the coil is varied through all possible positions, the maximum torque on the coil by magnetic forces is 0.16 N . m when the current in the coil is 4.0 mA.What is the magnitude of the magnetic field?

Accepted Answer

The torque on a current-carrying coil in a magnetic field is given by:
τ = NIABsinθ
where
N = number of turns in the coil
I = current in the coil
A = area of the coil
B = magnetic field strength
θ = angle between the normal to the coil and the direction of the magnetic field

In this case, the maximum torque is given as 0.16 Nm, N = 160, A = πr^2 = 0.16π m^2, I = 4.0 mA = 0.004 A. The angle θ is 90 degrees for maximum torque. Substituting these values, we get:
0.16 = (160)(0.004)(π)(0.4)^2B
B = 0.50 T

Therefore, the correct choice is C) 0.50 T.

Question 20

The figure shows the orientation of a rectangular loop consisting of 80 closely wrapped turns each carrying a current I.The magnetic field in the region is () mT.The loop can turn about the y axis.If $\theta$ = 30 $\degree$, a = 0.40 m, b = 0.30 m, and I = 8.0 A, what is the magnitude of the torque exerted on the loop?

Accepted Answer

The answer of The figure shows the orientation of a...

Quiz 8: Magnetic Forces

A positively charged particle has a velocity in the negative z direction at point P.The magnetic force on the particle at this point is in the negative y direction.Which one of the following statements about the magnetic field at point P can be determined from this data?

A straight wire is bent into the shape shown.Determine the net magnetic force on the wire when the current I travels in the direction shown in the magnetic field

An electron has a velocity of 6.0 *106 m/s in the positive x direction at a point where the magnetic field has the components Bx = 3.0 T, By = 1.5 T, and Bz = 2.0 T.What is the magnitude of the acceleration of the electron at this point?

A 2.0-m wire carries a current of 15 A directed along the positive x axis in a region where the magnetic field is uniform and given by B = (30 -40) mT.What is the resulting magnetic force on the wire?

An electron moving in the positive x direction experiences a magnetic force in the positive z direction.If Bx = 0, what is the direction of the magnetic field?

What current must be maintained in a square loop (50 cm on a side) to create a torque of 1.0 N . m about an axis through its centre and parallel to one of its sides when a magnetic field of magnitude 70 mT is directed at 40 $\degree$ to the plane of the loop?

A circular loop (radius = 0.50 m) carries a current of 3.0 A and has unit normal vector of () /3.What is the x component of the torque on this loop when it is placed in a uniform magnetic field of () T?

A 2.0-C charge moves with a velocity of () m/s and experiences a magnetic force of () N.The x component of the magnetic field is equal to zero.Determine the y component of the magnetic field.

A straight wire carries a current of 40 A in a uniform magnetic field (magnitude = 80 mT) .If the force per unit length on this wire is 2.0 N/m, determine the angle between the wire and the magnetic field.

What is the magnitude of the magnetic force on a charged particle (Q = 5.0 $\mu$ C) moving with a speed of 80 km/s in the positive x direction at a point where Bx = 5.0 T, By = -4.0 T, and Bz = 3.0 T?

A segment of wire carries a current of 25 A along the x axis from x = -2.0 m to x = 0 and then along the y axis from y = 0 to y = 3.0 m.In this region of space, the magnetic field is equal to 40 mT in the positive z direction.What is the magnitude of the force on this segment of wire?

A wire (mass = 50 g, length = 40 cm) is suspended horizontally by two vertical wires which conduct a current I = 8.0 A, as shown in the figure.The magnetic field in the region is into the paper and has a magnitude of 60 mT.What is the tension in either wire?

Quiz 8: Magnetic Forces

A positively charged particle has a velocity in the negative z direction at point P.The magnetic force on the particle at this point is in the negative y direction.Which one of the following statements about the magnetic field at point P can be determined from this data?

A straight wire is bent into the shape shown.Determine the net magnetic force on the wire when the current I travels in the direction shown in the magnetic field

An electron has a velocity of 6.0 *106 m/s in the positive x direction at a point where the magnetic field has the components Bx = 3.0 T, By = 1.5 T, and Bz = 2.0 T.What is the magnitude of the acceleration of the electron at this point?

A 2.0-m wire carries a current of 15 A directed along the positive x axis in a region where the magnetic field is uniform and given by B = (30 -40) mT.What is the resulting magnetic force on the wire?

An electron moving in the positive x direction experiences a magnetic force in the positive z direction.If Bx = 0, what is the direction of the magnetic field?

What current must be maintained in a square loop (50 cm on a side) to create a torque of 1.0 N . m about an axis through its centre and parallel to one of its sides when a magnetic field of magnitude 70 mT is directed at 40 °\degree° to the plane of the loop?

A circular loop (radius = 0.50 m) carries a current of 3.0 A and has unit normal vector of () /3.What is the x component of the torque on this loop when it is placed in a uniform magnetic field of () T?

A 2.0-C charge moves with a velocity of () m/s and experiences a magnetic force of () N.The x component of the magnetic field is equal to zero.Determine the y component of the magnetic field.

A straight wire carries a current of 40 A in a uniform magnetic field (magnitude = 80 mT) .If the force per unit length on this wire is 2.0 N/m, determine the angle between the wire and the magnetic field.

A particle (mass = 6.0 mg) moves with a speed of 4.0 km/s in a direction that makes an angle of 37 °\degree° above the positive x axis in the xy plane.At the instant it enters a magnetic field of (5.0) mT it experiences an acceleration of (8.0) m/s2.What is the charge of the particle?

What is the magnitude of the magnetic force on a charged particle (Q = 5.0 μ\muμ C) moving with a speed of 80 km/s in the positive x direction at a point where Bx = 5.0 T, By = -4.0 T, and Bz = 3.0 T?

A particle (mass = 2.0 mg, charge = -6.0 μ\muμ C) moves in the positive direction along the x axis with a velocity of 3.0 km/s.It enters a magnetic field of () mT.What is the acceleration of the particle?

A segment of wire carries a current of 25 A along the x axis from x = -2.0 m to x = 0 and then along the y axis from y = 0 to y = 3.0 m.In this region of space, the magnetic field is equal to 40 mT in the positive z direction.What is the magnitude of the force on this segment of wire?

A wire (mass = 50 g, length = 40 cm) is suspended horizontally by two vertical wires which conduct a current I = 8.0 A, as shown in the figure.The magnetic field in the region is into the paper and has a magnitude of 60 mT.What is the tension in either wire?

A uniform magnetic field of 0.50 T is directed along the positive x axis.A proton moving with a speed of 60 km/s enters this field.The helical path followed by the proton shown has a pitch of 5.0 mm.Determine the angle between the magnetic field and the velocity of the proton.

What current must be maintained in a square loop (50 cm on a side) to create a torque of 1.0 N . m about an axis through its centre and parallel to one of its sides when a magnetic field of magnitude 70 mT is directed at 40 $\degree$ to the plane of the loop?

What is the magnitude of the magnetic force on a charged particle (Q = 5.0 $\mu$ C) moving with a speed of 80 km/s in the positive x direction at a point where Bx = 5.0 T, By = -4.0 T, and Bz = 3.0 T?