Physics II Midterm Review
A 1.90 × 10¹⁴ Hz electromagnetic wave propagates in carbon tetrachloride with a speed of 2.05 × 10⁸ m/s. The wavelength of the wave in vacuum is closest to: A) 1,740 nm B) 1,580 nm C) 1,420 nm D) 1,080 nm E) 1,260 nm
1,580 nm Solution: don't get fooled by the velocity they give, because the question is asking for λ *in a vacuum*: λ = v/f λ = 3.0 × 10⁸/1.90 × 10¹⁴ λ = 1.579 × 10⁻⁶ λ = 1579 × 10⁻⁹ λ ≅ 1580 nm
A very long wire generates a magnetic field of 0.0020 × 10⁻⁴ T at a distance of 1.0 mm. What is the magnitude of the current? A) 2.0 mA B) 3,100 mA C) 1.0 mA D) 4,000 mA
1.0 mA Solution: B = μ₀I/2πr I = 2πrB/μ₀ I = 2(3.14)(1 × 10⁻³)(0.0020 × 10⁻⁴)/(4π × 10⁻⁷) I = 0.001 A or 1 × 10⁻³ A or 1.0 mA
The speed of light in a material is 0.41 c. What is the critical angle of a light ray at the interface between the material and a vacuum? A) 19° B) 22° C) 24° D) 17°
24° Solution: 0.41 × (3.0 × 10⁸) = 1.23 × 10⁸ n = c/v n = (3.0 × 10⁸)/(1.23 × 10⁸) n = 2.44 sinθc = n₂/n₁ sinθc = (1.00)/(2.44) θc = 24.19°
(*) A rigid rectangular loop, which measures 0.30 m by 0.40 m, carries a current of 9.9 A, as shown. A uniform external magnetic field of magnitude 1.8 T in the negative x-direction is present. Segment CD is in the x-z plane and forms a 19° angle with the z-axis, as shown. In Figure 10.5a, the y-component of the magnetic force on segment AB is closest to: A) -5.1 N B) zero C) +1.7 N D) +5.1 N E) -1.7 N
+5.1 N Solution: F = IlBsinθ I = 9.9 A l = 0.30 m since we are considering segment AB B = 1.8 T θ = 71° F = (9.9)(0.30)(1.8)sin71° F = 5.05 N, which is closest to +5.1 N.
At a certain instant in time, an electromagnetic wave has E in the -z direction and B in the +y direction. In what direction does the wave propagate? A) -z direction B) +x direction C) +y direction D) -x direction E) +z direction
+x direction
An erect object is 50 cm from a concave mirror of radius 60 cm. In Situation 24.1, the lateral magnification of the image is closest to: A) +1.5 B) -0.7 C) +0.4 D) +0.7 E) -1.5
-1.5
A circular loop of radius 10 cm and three long straight wires carry currents of I1 = 20 A, I2 = 50 A, I3 = 20 A, and I4 = 40 A, respectively, as shown. Each straight wire is 20 cm from the center of the loop. In Figure 20.8, the y-component of the resultant magnetic field at the center of the loop is closest to: A) -50 μT B) -60 μT C) +60 μT D) -54 μT E) +50 μT
-54 μT Explanation: the magnetic fields due to I₁ and I₂ are in the z direction, so they can be ignored B₁ = μ₀I₃/2πr₁ B₁ = (4π * 10⁻⁷ T•m/A(20A)/(2)(3.14)(20 × 10⁻²) B₁ = 2 × 10⁻⁵ B₂ = μ₀I₄/2πr₂ B₂ =.(4π * 10⁻⁷ T•m/A(40A)/(2)(3.14)(20 × 10⁻²) B₁ = 4 × 10⁻⁵ For B₁, must multiply by the cos45°. 2 × 10⁻⁵ • √2/2 = 1.414 × 10⁻⁵ Then add them together: 1.414 × 10⁻⁵ + 4 × 10⁻⁵ = 5.414 × 10⁻⁵ or 54 × 10⁻⁶ T or 54 μT
A uniform magnetic field of magnitude 0.80 T in the negative z-direction is present in a region of space. A uniform electric field is also present. In Figure 20.3, the electric field is set at 20,800 V/m in the positive y-direction. An electron is projected with an initial velocity v₀ = 2.6 × 10⁴ m/s in the positive x-direction. The y-component of the initial force on the electron is closest to: A) zero B) -3 × 10⁻¹⁵ N C) +7 × 10⁻¹⁵ N D) +3 × 10⁻¹⁵ N E) -7 × 10⁻¹⁵ N
-7 × 10⁻¹⁵ N
A sinusoidal electromagnetic wave is propagating in vacuum. At a given point and at a particular time the electric field is in the +x direction and the magnetic field is in the -y direction. What is the direction of propagation of the wave?
-z direction (into the page)
A charged particle of mass 0.0040 kg is subjected to a 4.0 T magnetic field which acts at a right angle to its motion. If the particle moves in a circle of radius 0.10 m at a speed of 2.0 m/s, what is the magnitude of the charge on the particle? A) 2,500 C B) 0.020 C C) 0.00040 C D) 50 C
0.020 C Solution: qB = mv/r q = mv/rB q = (0.0040)(2)/(0.10)(4.0) q = 0.020 C
(*) A radiometer has two square vanes (1 cm by 1 cm), attached to a light horizontal cross arm, and pivoted about a vertical axis through the center. The center of each vane is 6 cm from the axis. One vane is silvered and it reflects all radiant energy incident upon it. The other vane is blackened and it absorbs all incident radiant energy. Radiant energy, which has an intensity of 300 W/m²0.03 , is incident normally upon the vanes. In Figure 23.1, the radiant power absorbed by the blackened vane is closest to: A) 0.04 W B) 0.06 W C) 0.09 W D) 0.03 W E) 0.05 W
0.03 W Solution: A = (1 cm)(1 cm) = (1 × 10⁻² m)(1 × 10⁻² m) = 1 × 10⁻⁴ m² P = IA P = (300)(1 × 10⁻⁴) = 300 × 10⁻⁴ = 0.03
(*) Alpha particles (charge = +2e, mass = 6.68 × 10-27 kg) are accelerated in a cyclotron 1) to a final orbit radius of 0.30 m. The magnetic field in the cyclotron is 0.80 T. The period of circular motion of the alpha particles is closest to: A) 0.33 μs B) 0.49 μs C) 0.40 μs D) 0.25 μs E) 0.16 μs
0.16 μs
A 5 meter length of wire carrying a current of 5 A lies on a horizontal table with a rectangular top of dimensions 0.300 m × 0.400 m. The ends of the wire are attached to opposite ends of a diagonal of the rectangle. A vertical magnetic field of 0.30 T is present. What magnetic force acts on this segment of wire? A) 0.75 N B) 7.5 N C) zero D) 1.1 N E) The force cannot be determined without knowing the shape of the length of wire.
0.75 N Solution: To find the diagonal: L = √ [(0.30 m)2 + (0.40 m)2] = 0.50 m F = IlB F = (5)(0.5)(0.30) F = 0.75 N
A rigid rectangular loop, which measures 0.30 m × 0.40 m, carries a current of 7.7 A, as shown. A uniform external magnetic field of maggitude 2.6 T in the negative x-direction is present. Segment CD is in the x-z plane and forms a 33° angle with the z-axis, as shown. In Figure 20.5c, the magnitude of the magnetic moment of the loops is closest to: A) 0.31 A∙m² B) 0.46 A∙m² C) 0.62 A∙m² D) 0.92 A∙m² E) 0.77 A∙m²
0.92 A•m² μ = IA μ = (7.7)(0.12) μ = 0.92 A•m²
In Figure 23.1, the radiation pressure on the blackened vane is closest to: A) 1 × 10⁻¹⁰ Pa B) 1 × 10⁻⁷ Pa C) 1 × 10⁻⁶ Pa D) 1 × 10⁻⁹ Pa E) 1 × 10⁻⁸ Pa
1 × 10⁻⁶ Pa Solution: P = F/A = I/c P = 300/3.0 × 10⁸ P = (3 × 10²)/(3 × 10⁸) P = 1 × 10⁻⁶
A mass spectrograph is operated with deuterons, which have a charge of +e and a mass of 3.34 × 10⁻²⁷ kg. Deuterons emerge from the source, which is grounded with negligible velocity. The velocity of the deuterons as they pass through the accelerator grid is 8.0 × 105 m/s. A uniform magnetic field of magnitude B = 0.20 T, directed out of the plane, is present at the right of the grid. In Figure 20.4, the angular velocity of the deuterons in the magnetic field is closest to: A) 1.6 × 10⁶ rad/s B) 6.3 × 10⁷ rad/s C) 4.0 × 10⁶ rad/s D) 2.5 × 10⁷ rad/s E) 1.0 × 10⁷ rad/s
1.0 × 10⁷ rad/s Solution: ω = v/r we know v and r was found in the previous problem: 80 mm ω = 8.0 × 10⁵/8.0 × 10⁻² ω = 1.0 × 10⁷ rad/s
A rigid rectangular loop, which measures 0.30 m by 0.40 m, carries a current of 5.5 A, as shown. A uniform external magnetic field of magnitude 2.9 T in the negative x-direction is present. Segment CD is in the x-z plane and forms a 35° angle with the z-axis, as shown. In Figure 20.5b, an external torque applied to the loop keeps it in static equilibrium. The magnitude of the external torque is closest to: A) 0.73 N∙m B) 1.4 N∙m C) 1.3 N∙m D) 1.6 N∙m E) 1.1 N∙m
1.1 N•m Solution: τ = nBIAsinθ A = (0.30)(0.40) = 0.12 m² τ = (1)(2.9)(5.5)(0.12)sin35° τ = 1.0978, which is closest to 1.1 N•m
A glass plate whose index of refraction is 1.66 is immersed in a liquid. The surface of 27) the glass is inclined at an angle of 44° with the vertical. A horizontal ray in the glass is incident on the interface. When the liquid is a certain alcohol, the incident ray arrives at the interface at the critical angle. The index of refraction of the alcohol is closest to: A) 1.15 B) 1.17 C) 1.09 D) 1.11 E) 1.13
1.15
(*) A double convex thin glass lens has equal radii of curvature. The focal length of the lens is +31.2 cm and the index of refraction of the glass is 1.52. The lens is replaced with a plano-convex glass lens of the same focal length and thickness. The radius of curvature of the convex surface is 13.0 cm. The index of refraction of the glass of the plano-convex lens is closest to: A) 1.38 B) 1.42 C) 1.36 D) 1.40 E) 1.44
1.42 R₁ = +13 R₂ = ∞ 1/f = (n - 1)(1/R₁ - 1/R₂) 1/31.2 = (n - 1)(1/13 - 1/∞) 0.0321 = (n - 1)(0.0769 - 0) 0.417 = n - 1 n = 1.417
Light having a speed in vacuum of 3.0 × 108 m/s enters a liquid of refractive index 2.0. In this liquid, its speed will be A) 3.0 × 108 m/s B) 1.5 × 108 m/s C) 0.75 × 108 m/s D) 6.0 × 108 m/s E) None of the above choices are correct.
1.5 × 10⁸ m/s
A solenoid of length 10 cm consists of a wire wrapped tightly around a wooden core. The magnetic field strength is 4.0 T inside the solenoid. If the solenoid is stretched to 25 cm by applying a force to it, what does the magnetic field become? A) 4.0 T B) 10.0 T C) 1.6 T D) 20 T
1.6 T
An electron enters a magnetic field of 0.69 T with a velocity perpendicular to the direction of the field. At what frequency does the electron traverse a circular path? (The mass of an electron is 9.1 × 10⁻³¹ kg, and the charge of an electron is 1.6 × 10⁻¹⁹ C.) A) 1.9×10¹⁰ Hz B) 5.3×10⁻¹¹ Hz C) 1.9×10¹⁴ Hz D) 5.3×10⁻⁷ Hz
1.9×10¹⁰ Hz
The critical angle for an air-glass interface is 10.9 degrees. A ray in air is incident on the interface. The reflected ray is 100 percent polarized. The angle of refraction is closest to: A) 10.1° B) 8.91° C) 9.51° D) 8.31° E) 10.7°
10.7°
(*) A fish appears to be 9.00 m below the surface of a pond when viewed almost directly above by a fisherman. What is the actual depth of the fish? (nwater = 1.33) A) 0.08 m B) 6.77 m C) 0.15 m D) 11.97 m
11.97 cm Solution: n water = real depth/apparent depth 1.33 = real depth/9.00 real depth = 11.97 cm
An electron in a magnetic field has a cyclotron frequency of 3.0 x 10¹² Hz. What is the magnetic field strength? (The mass of an electron is 9.1 x 10⁻³¹ kg, and the charge of an electron is 1.6 x 10⁻¹⁹ C.) A) 12 T B) 0.0093 T C) 0.084 T D) 110 T
110 T
The energy flow per unit time per unit area (S) of an electromagnetic wave has an average value of 695 mW/m². The wave is incident upon a rectangular area, 1.5 m by 2.0 m, at right angles. The total energy that traverses the area in a time interval of one minute is closest to: A) 190 J B) 130 J C) 160 J D) 220 J E) 250 J
130 J Solution: Work = power × time Power = intensity × area Work = (intensity × area) × time Work = (695 × 10⁻³)(3)(60) Work = 125 J, which is closest to 130 J
An erect object is 50 cm from a concave mirror of radius 60 cm. In Situation 24.1, the object is moved to a new position, such that the new lateral 16) magnification is +2.5. The new object distance, in cm, is closest to: A) 18 B) 24 C) 30 D) 36 E) 42
18
A tank holds a layer of oil, 1.95 m thick, which floats on a layer of syrup that is 0.64 m thick. Both liquids are clear and do not intermix. A ray, which originates at the bottom of the tank on a vertical axis, crosses the oil-syrup interface at a point 0.90 m from the axis. The ray continues and arrives at the oil-air interface, 2.00 m from the axis and at the critical angle. In Figure 23.3a, the index of refraction of the oil is closest to: A) 2.06 B) 2.00 C) 2.04 D) 2.08 E) 2.02
2.04 Solution: Horizontal displacement = 2.0m - 0.9m = 1.1m The Vertical displacement is the thickness of the oil, which is 1.95m. The angle of this beam is: tan⁻¹( 1.1m/1.95m ) = 29.43º This is the critical angle for the oil/air interface. θc = 29.43º. θc = sin⁻¹(n air)/(n oil) θc = sin⁻¹(1/(n oil) sinθc = sin[sin⁻¹(1/(n oil) ] sinθc = 1/(n oil) sin(29.43º) = 1/(n oil) 0.4914 = 1/(n oil) n oil = 1/0.4914 index of oil = 2.035
An 800 kHz radio signal is detected at a point 9.5 km distant from a transmitter tower. The electric field amplitude of the signal at that point is 230 mV/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. The average electromagnetic energy density at that point is closest to: A) 4.7 × 10⁻¹³ J/m³ B) 6.6 × 10⁻¹³ J/m³ C) 3.3 × 10⁻¹³ J/m³ D) 9.4 × 10⁻¹³ J/m³ E) 2.3 × 10⁻¹³ J/m³
2.3 × 10⁻¹³ J/m³
An 800 kHz radio signal is detected at a point 6.6 km distant from a transmitter tower. The electric field amplitude of the signal at that point is 780 mV/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. The magnetic field amplitude of the signal at that point, in nT, is closest to: A) 3.1 B) 3.6 C) 2.6 D) 1.6 E) 2.1
2.6
(*) A 7.55 × 10¹⁴ Hz electromagnetic wave propagates in carbon tetrachloride with a speed of 2.05 × 10⁸ m/s. The wavelength of the wave in carbon tetrachloride is closest to: A) 361 nm B) 397 nm C) 272 nm D) 301 nm E) 338 nm
272 nm Solution: v = λf λ = v/f = 2.05 × 10⁸/7.55 × 10¹⁴ = 2.72 × 10⁻⁷ m or 272 × 10⁻⁹ m or 272 nm
When unpolarized light from air (n = 1) is incident on a piece of glass with index of refraction n = 1.80, the reflected light is found to be completely polarized when the angle of incidence is θB. What is the angle of refraction in this case? Give a numerical answer.
29.1° *When light strikes a surface at the polarizing angle, the reflected and refracted rays are *perpendicular* to each other. In this case, the angle of refraction θb is equal to 90° - θ and tanθp = nb/na.
Alpha particles (charge = +2e, mass = 6.68 × 10-27 kg) are accelerated in a cyclotron to a final orbit radius of 0.50 m. The magnetic field in the cyclotron is 0.50 T. The kinetic energy of an alpha particle in the final orbit is closest to: A) 3.0 MeV B) 2.6 MeV C) 4.3 MeV D) 3.9 MeV E) 3.4 MeV
3.0 MeV last step: convert J to MeV 1 eV = 1.602 × 10⁻¹⁹ J So, 0.048 × 10⁻¹¹ J × 1eV/1.602 × 10⁻¹⁹ J = 0.03 × 10⁸ eV 1 MeV = 1 × 10⁶ eV So, 3.0 × 10⁶ eV × 1MeV/1 × 10⁶ eV = 3.0 MeV Final answer: 3.0 MeV
A circular coil of wire of 200 turns and diameter 6 cm carries a current of 7 A. It is 14) placed in a magnetic field of 0.90 T with the plane of the coil making an angle of 30° with the magnetic field. What is the torque on the coil? A) 1.8 N∙m B) 5.9 N∙m C) 3.1 N∙m D) 8.2 N∙m E) 1.5 N∙m
3.1 N∙m Solution: τ = nBIAsinθ A = π(0.03)² = 0.0028 θ = 60° made by the normal to the plane of the loop τ = (200)(0.90)(7)(0.0028)sin60° τ = 3.08 N∙m, which is closest to 3.1 N∙m
A square wire 2.0 cm on each side is oriented so that a 5.00 mT magnetic field makes a 18) 30.0° angle with the normal to the plane of the square. If a 1.0 mA current causes a 32.0 N∙m torque on the wire, how many loops does the wire make? A) 1.6 × 10¹⁰ B) 3.2 × 10⁶ C) 3.2 × 10¹⁰ D) 4.5 × 10¹⁰
3.2 × 10¹⁰ Solution: τ = nBIAsinθ 32 = n(5 × 10⁻³)(1 × 10⁻³)(4.0 × 10⁻⁴)sin30° 32 = n(10 × 10⁻¹⁰) n = 3.2 × 10¹⁰ turns
A 2.4 x 10¹⁴ Hz laser emits a 3.5 μs pulse, 5.0 mm in diameter, with a beam energy density of 0.65 J/m³. The power emitted by the laser is closest to: A) 3.8 kW B) 12 kW C) 7.7 kW D) 19 kW E) 15 kW
3.8 kW
The magnitude of a magnetic field a distance 3.0 μm from a wire is 20.0 × 10⁻⁴ T. How much current is flowing through the wire. Assume the wire is the only contributor to the magnetic field. A) 30 mA B) 377 mA C) 19 mA D) 188 mA
30 mA
Light having a wavelength in vacuum of 600 nm enters a liquid of refractive index 2.0. In this liquid, its wavelength will be A) 300 nm B) 600 nm C) 1200 nm D) 150 nm E) None of the above choices are correct.
300 nm
What is the wavelength used by a radio station that broadcasts with a carrier frequency of 920 kHz? A) 22.6 m B) 276 m C) 226 m D) 326 m E) 175 m
326 m Solution: λ = v/f λ = 3.0 × 10⁸/(920 × 10³) λ = 0.00326 × 10⁵ λ = 326 m
A ray in glass is incident onto a water-glass interface, at an angle of incidence equal to half the critical angle for that interface. The indices of refraction for water and the glass are 1.33 and 1.43, respectively. The angle that the refracted ray in the water makes with the normal is closest to: A) 42° B) 32° C) 47° D) 27° E) 37°
37°
A double convex thin glass lens has equal radii of curvature. The focal length of the lens is +37.3 cm and the index of refraction of the glass is 1.52. The radius of curvature of each convex surface, in cm, is closest to: A) 46 B) 31 C) 35 D) 39 E) 43
39
Two parallel, straight wires are 7.0 cm apart and each carries a 18.0 A current in the same direction. One wire is securely anchored, and the other is attached in the center to a movable cart. If the force needed to move the wire when it is not attached to the cart is negligible, with what force does the wire pull on the cart? Both wires are 50.0 m long. A) 9,300 dynes B) 660 dynes C) 370 dynes D) 4,600 dynes
4,600 dynes Solution: 1 dyne = 1 × 10⁻⁵ N So 0.0463 N × 1 dyne/1 × 10⁻⁵ N = 4,630 dynes, which is closest to 4,600 dynes.
A certain electromagnetic field traveling in vacuum has a maximum electric field of 1200 V/m. What is the maximum magnetic field of this wave? A) 2.2 × 10⁻⁵ T B) 4.0 × 10⁻⁶ T C) 8.7 × 10⁻⁶ T D) 9.6 × 10⁻⁶ T E) 3.4 × 10⁻⁴ T
4.0 × 10⁻⁶ T Solution: E = cB B = E/c B = 1200/3.0 × 10⁸ B = 4.0 × 10⁻⁶
A solenoid 3.0 cm long consists of 5,748 loops of wire. If the magnetic field inside the solenoid is 1.0 T, what is the magnitude of the current that flows through it? A) 3.0 A B) 52 A C) 0.24 A D) 4.2 A
4.2 A Solution: For a solenoid, the field inside is B = μ₀nI where n is the number of turns per meter. In this case, n = (5748)/(.03 m) = 1.91 × 10⁵ m⁻¹ Solving for I, I = B/μ₀n = (1.0 T)/((4π * 10⁻⁷ T•m/A)(1.91 × 10⁵ m⁻¹)) = 4.2 A
In Figure 23.5, in the investigation of a new type of optical fiber (index of refraction n = 1.19), a laser beam is incident on the flat end of a straight fiber in air. What is the maximum angle of incidence θ₁ if the beam is not to escape from the fiber? A) 37.1° B) 33.9° C) 27.7° D) 40.2° E) 30.8°
40.2° Solution: For internal reflection inside the optical fibre... 1.19sinθᵢ = 1sin90° sinθᵢ = 1/1.19 sinθᵢ = 0.840 θᵢ = 57.18° Now consider the initial air-fiber interface: Here the angle of refraction will be (90 - 57.18) = 32.82° So, 1sinθ₁ = 1.19sin32.82° θ₁ = sin⁻¹(0.645) θ₁ = 40.16°
(*K*) An amateur astronomer grinds a double convex lens whose surfaces have radii of curvature of 40 cm and 60 cm. The glass has an index of refraction of 1.54. What is the focal length of this lens in air? A) 42.5 cm B) 44.4 cm C) 126 cm D) 222 cm E) 88.8 cm
44.4
A ray of light traveling in air makes a 63.0° angle with respect to the normal of the surface of a liquid. It travels in the liquid at a 39.2° angle with respect to the normal. Find the critical angle for total internal reflection. (Assume n = 1 for air.)
45.2°
A cylindrical tank is 50.0 ft deep, 31.0 ft in diameter, and filled to the top with water. A 31) flashlight shines into the tank from above. What minimum angle φ can its beam make with the water surface if the beam is to illuminate part of the bottom? (nwater = 1.33) A) 41.1° B) 49.6° C) 45.5° D) 53.5°
45.5° Solution: tan⁻¹(31/50) = 31.79° n₁sinθ₁ = n₂sinθ₂ 1sinθ₁ = 1.33sin31.79° θ₁ = 44.5° θᵢ = 90 - θ₁ θᵢ = 45.5°
In Figure 20.2 is a velocity selector that can be used to measure the speed of a charged particle. A beam of particles is directed along the axis of the instrument. A parallel plate capacitor sets up an electric field E which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by 3 mm and the value of the magnetic field is 0.3 T, what voltage between the plates will allow particles of speed 5 x 10⁵ m/s to pass straight through without deflection? A) 70 V B) 1,400 V C) 450 V D) 2,800 V E) 140 V
450 V Solution: Because the electric field and magnetic field are applied simultaneously: F electric = F magnetic qE = qvB E = vB So, v = E/B Apply v = E/B 5 × 10⁵ m/s = E/0.3 T E = 1.5 × 10⁵ N/C Then use V = Ed V = (1.5 × 10⁵)(3 × 10⁻³) V = 4.5 × 10² Final answer: 450 V
A ray in air is incident on a glass plate whose index of refraction is 1.42. The angle of refraction is one half the angle of reflection. The angle of refraction is closest to: A) 39° B) 45° C) 41° D) 43° E) 37°
45°
A very long straight current-carrying wire produces a magnetic field of 20 mT at a distance d from the wire. To measure a field of 5 mT due to this wire, you would have to go to a distance of: A) 8d B) d√2 C) 4d D) 16d E) 2d
4d Solution: B₁ = μ₀I/2πd 20 mT = μ₀I/2πd B₂ = μ₀I/2πd 5 mT = μ₀I/2πd₂ 20 mT = μ₀I/2πd ------------------ 5 mT = μ₀I/2πd₂ 4 mT = d₂/d d₂ = 4d
The radius of curvature of the curved side of a plano-convex lens made of glass 8) (n = 1.64) is 33 cm. What is the focal length of the lens? A) 21 cm B) -21 cm C) 52 cm D) -52 cm
52 cm
In Figure 23.7, a ray of light in air is incident on a flat piece of glass, at an angle of φ₀ = 84° with respect to the normal. The glass has an index of refraction n = 1.5. What is the angle θ between the reflected (ray b) and refracted (ray c) rays?
54°
Light having a frequency in vacuum of 6.0 × 10¹⁴ Hz enters a liquid of refractive index 2.0. In this liquid, its frequency will be: A) 1.5 × 10¹⁴ Hz B) 12 × 10¹⁴ Hz C) 6.0 × 10¹⁴ Hz D) 3.0 × 10¹⁴ Hz E) None of the above choices are correct.
6.0 × 10¹⁴ Hz Solution: frequency of light does *not* change passing from one medium to another
An object is placed 15.0 cm to the left of a double-convex lens of focal length 20.0 cm. The image of this object is located A) 60.0 cm to the right of the lens. B) 8.57 cm to the right of the lens. C) 60.0 cm to the left of the lens. D) 8.57 cm to the left of the lens. E) 30.0 cm to the left of the lens.
60.0 cm to the left of the lens.
In Figure 23.4, a laser positioned on a ship is used to communicate with a small two man research submarine resting on the bottom of a lake. The laser is positioned 12 meters above the surface of the water, and it strikes the water 20 meters from the side of the ship. The water is 58 meters deep and has an index of refraction of 1.33. How far is the submarine from the side of the ship? A) 48.9 m B) 88.9 m C) 78.9 m D) 58.9 m E) 68.9 m
68.9
A wire carrying a current is shaped in the form of a circular loop of radius 4.0 mm. If the magnetic field strength at its center is 1.1 mT with no external magnetic fields contributing to it, what is the magnitude of the current that flows through the wire? A) 28 A B) 14 A C) 17 A D) 7.0 A
7.0 A
(*) A mass spectrograph is operated with deuterons, which have a charge of +e and a mass of 3.34 × 10⁻²⁷ kg. Deuterons emerge from the source, which is grounded with negligible velocity. The velocity of the deuterons as they pass through the accelerator grid is 8.0 × 105 m/s. A uniform magnetic field of magnitude B = 0.20 T, directed out of the plane, is present at the right of the grid. In Figure 20.4, the deuterons are in circular orbit in the magnetic field. The radius of the orbit and the initial sense of deflection are closest to: A) 70 mm, upward B) 80 mm, downward C) 70 mm, downward D) 60 mm, upward E) 60 mm, downward
80 mm, downward Solution: v = qBr/m r = mv/qB r = (3.34 × 10⁻²⁷)(8.0 × 10⁵)/(1.6 × 10⁻¹⁹)(0.20) r = 83.5 × 10⁻³ m or 83 mm thus, closest to B. Using the right hand rule: arm points right, fingers point out of the page, thumb naturally points *downwards* then
Which statement about thin lenses is correct? In each case, we are considering only a single lens. A) A converging lens always produces a real inverted image. B) A diverging lens always produces a virtual inverted image. C) A converging lens sometimes produces a real erect image. D) A diverging lens produces a virtual erect image only if the object is located within the focal point of the lens. E) A diverging lens always produces a virtual erect image.
A diverging lens always produces a virtual erect image.
Which statement about images is correct? A) A virtual image cannot be formed on a screen. B) A virtual image cannot be photographed. C) A virtual image cannot be viewed by the unaided eye. D) Mirrors always produce real images because they reflect light. E) A real image must be erect.
A virtual image cannot be formed on a screen
(*) Which type of electromagnetic wave travels through space the slowest? A) All EM waves travel at the same speed. B) microwaves C) ultraviolet D) radio waves E) infrared
All EM waves travel at the same speed
A wire along the z-axis carries a current of 4.9 A in the positive z direction. Find 38) the force (magnitude and direction) exerted on a 3.3 cm long length of the wire by a uniform magnetic field with magnitude 0.43 T in the -x direction.
F = IlBsinθ F = (4.9)(3.3 × 10⁻²)(0.43)(1) F = 6.95 × 10⁻² N ≅ 7 × 10⁻² N or 0.070 N for the magnitude As for the direction, it must be perpendicular to both I and B, so in the y direction. Using the right hand rule, arm points towards my body, fingers point to the left, thumb points naturally down. Final answer: 0.070 N, -y direction
(*) A ray in glass arrives at the glass-water interface at an angle of 48° with the normal. The refracted ray, in water, makes a 63° angle with the normal. The index of refraction of water is 1.33. In Figure 23.2a, the angle of incidence of a different ray in the glass is 11°. The angle of refraction in the water is closest to: A) 13° B) 17° C) 9.2° D) 11° E) 15°
I HAVE 13° ASK ABOUT THIS ONE
Suppose you place your face in front of a concave mirror. A)Your image will always be inverted. B) Your image will be diminished in size. C) If you position yourself between the center of curvature and the focal point of the mirror, you will not be able to see your image. D) No matter where you place yourself, a real image will be formed. E) None of these is true.
If you position yourself between the center of curvature and the focal point of the mirror, you will not be able to see your image. Explanation: because the image will be formed behind you, so you cannot see the image if it placed between focal length and radius of curvature.
A proton, with mass 1.67 × 10-27 kg and charge +1.6 × 10⁻¹⁹ C, is sent with velocity 2.3 × 10⁴ m/s in the x-direction into a region where there is a uniform electric field of magnitude 780 V/m in the y direction. What is the magnitude and direction of the uniform magnetic field in the region, if the proton is to pass through undeflected? Assume that the magnetic field has no x-component. Neglect gravitational effects.
Since both E and B are being generated simultaneously, Fe = Fb qE = qvB E = vB 780 = (2.3 × 10⁴)B B = 339 × 10⁻⁴ ≅ 3.4 × 10⁻² As for the direction, it must be perpendicular to both v and E, so in the z direction. Using the right hand rule, arm points right, fingers point up, thumb points naturally as if going out of the page (towards my body). Final answer: 3.4 × 10⁻² T, +z direction
If a single lens forms a virtual image of an object, then A) The lens could be either a diverging or a converging lens. B) The lens must be a diverging lens. C) The lens must be a converging lens. D) The image must be inverted.
The lens could be either a diverging or a converging lens.
A circular loop of radius 10 cm and three long straight wires carry currents of I1 = 60 A, I2 = 20 A, I3 = 70 A, and I4 = 30 A, respectively, as shown. Each straight wire is 20 cm from the center of the loop. In Figure 20.8, the z-component of the resultant magnetic field at the center of the loop is closest to: A) +40 μT B) -170 μT C) +360 D) -360 μT E) -40 μT
The magnetic fields due to I₃ and I₄ are in the x-y direction . The only contributionto the magnetic field in the z- direction is coming from I₂ and I₁. For I₁ Bc = (μ₀I₁/2R) (At the center of circular loop, directed in the -z) Bc = (4πx10⁻⁷)(60A)/2(0.1m) Bc = 377 μT ................................................. For I₂ B2 = (μ₀I₂/2πr) B2 = (4πx10⁻⁷)(20A)/2(0.2m) B2 = 20 μT ..........................................................the z-component of the resultant magnetic field at the center of the loop is closest to: B = 20 μT - 377 μT B = -357μT
(*) Which of the following is an accurate statement? A) The magnetic force on a moving charge does not change its energy. B) Magnetic field lines have as their sources north and south poles. C) A magnetic field line is, by definition, tangent to the direction of the magnetic force on a moving charge at a given point in space. D) The magnetic force on a current carrying wire is greatest when the wire is parallel to the magnetic field. E) A current carrying loop of wire tends to line up with its plane parallel to an external magnetic field in which it is positioned.
The magnetic force on a moving charge does not change its energy.
What is the essential difference between microwaves and blue light? A) Blue light is a beam of photons. Microwaves are not photons. B) One has an electric charge, the other does not. C) One undergoes refraction, the other does not. D) One is a form of radiation, the other is not. E) There is no essential difference in the nature of microwaves and blue light other than a difference in frequency and wavelength.
There is no essential difference in the nature of microwaves and blue light other than a difference in frequency and wavelength.
In Figure 20.14, the two long straight wires are separated by a distance of d = 0.40 m. The currents are I1 = 1.0 A to the right in the upper wire and I2 = 8.0 A to the left in the lower wire. What is the magnitude and direction of the magnetic field at point P, that is a distance d/2 = 0.20 m below the lower wire?
To find the magnetic field at the point you take the sum of the magnetic fields of each wire generated at that point. First start off with the bottom wire using the RHR you can figure out that the magnetic field should be out of the screen. Next take the equation B = (μ₀I)/(2π(d/2)) B = (4π × 10⁻⁷)(8.0)/(2)(3.14)(0.20) B = 8 × 10⁻⁶ T out of the field. Next you find the magnetic field created by the upper wire which is using the RHR directed into the screen, then you take the same formula as the lower wire except now it is B=(μ₀I)/(2π(1.5d)) B = (4π × 10⁻⁷)(1.0)/(2)(3.14)(0.6) B = 3.33 × 10⁻⁷ T into the screen Then you add them together taking into the screen to be negative so you get Btot = (8 × 10⁻⁶)-(3.33 × 10⁻⁷) = 7.67 × 10⁻⁶ T Final answer: 7.7 × 10⁻⁶ T directed out of the plane of the paper.
An object 3.4 mm tall is placed 25 cm from the vertex of a convex spherical mirror. The radius of curvature of the mirror has magnitude 73 cm. (a) How far is the image from the vertex of the mirror? (b) What is the height of the image?
a) 15 cm b) 2.0 mm
The left-hand end of a glass rod is ground to a spherical surface. The glass has index of refraction 1.50. A small object 4.00 mm tall is placed in the axis of the rod, 11.0 cm to the left of the vertex of the spherical surface. The image is formed in the rod, 17.0 cm to the right of the vertex. (a) What is the magnitude of the radius of curvature of the spherical surface at the end of the rod? (b) What is the height of the image?
a) 2.79 cm b) 4.12 mm
When an object is placed 118 cm from a diverging thin lens, its image is found to be 59 cm from the lens. The lens is removed, and replaced by a thin converging lens whose focal length is the same in absolute value as the diverging lens. This second lens is at the original position of the first lens. Where is the image of the object now?
at infinity Explanation: image could form fine with a diverging lens because those always produce a virtual image no matter where the object is. However, once you get to converging lens, if the object is placed at the focal point, no image will be formed (at ∞). That's exactly what we have calculated: f is 118 cm, and it was stated originally that the object distance was 118 cm.
Consider a solenoid of length L, N windings, and radius b (L is much longer than b). A current I is flowing through the wire. If the length of the solenoid became twice as long (2L), and all other quantities remained the same, the magnetic field inside the solenoid would: A) become half as strong as initially. B) stay the same. C) become twice as strong as initially.
become half as strong as initially
If you were to cut a small permanent bar magnet in half: A) one piece would be a magnetic north pole and the other piece would be a south pole. B) each piece would in itself be a smaller bar magnet with both north and south poles. C) neither piece would be magnetic. D) None of these statements is true.
each piece would in itself be a smaller bar magnet with both north and south poles
Two electromagnetic waves are traveling at the same speed. The wave with the higher wavelength: A) has a higher frequency than the other wave. B) is traveling slower than the other wave. C) is traveling faster than the other wave. D) has a lower frequency than the other wave.
has a lower frequency than the other wave.
A positive charge in Figure 20.1 is moving to the right and experiences a vertical (upward) magnetic force. In which direction is the magnetic field? A) out of the page B) into the page C) upward D) to the right E) to the left
into the page Explanation: arm to the right, thumb up, fingers point into the page.
(*) As you walk away from a plane mirror on a wall, your image A) is always a real image, no matter how far you are from the mirror. B) is always the same size. C) gets smaller. D) may or may not get smaller, depending on where the observer is positioned. E) changes from being a virtual image to a real image as you pass the focal point.
is always the same size.
When light travels from air into water, A) its velocity changes, but its frequency and wavelength do not change. B) its velocity and wavelength change, but its frequency does not change. C) its velocity, wavelength and frequency all change. D) its frequency changes, but its velocity and wavelength do not change. E) its wavelength changes, but its velocity and frequency do not change.
its velocity and wavelength change, but its frequency does not change.
A charged particle moving within a static magnetic field: A) will always experience a magnetic force, regardless of its direction of motion. B) may experience a magnetic force, but its speed will not change. C) may experience a magnetic force which will cause its speed to change. D) None of the above statements are true.
may experience a magnetic force, but its speed will not change. Explanation: A charged particle moving in a magnetic field experiences no force if it is parallel to the field. Magnetic forces are always perpendicular to the direction of motion, hence they do no work on particle. Therefore speed of particle remains the same.
In Figure 20.12, a small particle of charge q = -1.9 x 10⁻⁶ C and mass m = 3.1 × 10⁻¹² kg has velocity ν₀ = 8.1 × 10³ m/s as it enters a region of uniform magnetic field. The particle is observed to travel in the semicircular path shown, with radius R = 5.0 cm. Calculate the: (a) magnitude and (b) direction of the magnetic field in the region.
mv²/r = qvB mv/r = qB B = mv/rq B = (3.1 × 10⁻¹²)(8.1 × 10³)/(5 × 10⁻²)(-1.9 × 10⁻⁶) B = -2.64 × 10⁻¹ = -0.26 a) | B | = 0.26 b) Velocity vector is always tangent to the circle, so force is pointing inwards. Thumb naturally points out of the page, but this is a negatively charged particle, so it points into the page Final answer: a) 0.26 T b) directed into the paper
(*) An erect object is 50 cm from a concave mirror of radius 60 cm. In Situation 24.1, the character of the image is: A) indeterminate B) virtual and inverted C) real and erect D) real and inverted E) virtual and erect
real and inverted
A straight bar magnet is initially 4 cm long, with the north pole on the right and the south pole on the left. If you cut the magnet in half, the right half will: A) only contain a south pole. B) only contain a north pole. C) still contain a north pole on the right and a new south pole on the left.
still contain a North Pole on the right and a new South Pole on the left.
A metallic weight is suspended from a metal spring. If now a current is passed through the spring, A) the spring will extend, lowering the weight. B) the weight will not move. C) the spring will contract, raising the weight. D) whether or not the weight moves up or down depends on what the weight is made of (i.e. whether or not it is magnetizable). E) None of these are true.
the spring will contract, raising the weight.
A ring with a clockwise current (as seen from above the ring) is situated with its center directly above another ring, which has a counter-clockwise current, as shown in Figure 20.7. In what direction is the net magnetic force exerted on the top ring? A) upward C) The net force is zero. B) downward D) Impossible to determine
upward Solution: For the upper ring: According to Ampere's right hand rule, the current flows in the direction of the curled fingers of the right hand, and the magnetic field points in the downward direction of the ring. For the lower ring: According to Ampere's right hand rule, the current flows in the direction of the curled fingers of the right hand, and the magnetic field points in the upward direction of the ring. Parallel wires carrying current in the same direction attract. Parallel wires carrying current in opposite direction repel. Since these wires are carrying current in opposite directions, the rings will repel: The top ring will experience an upward force (away from the bottom ring).
(*) A negatively charged particle is moving to the right, directly above a wire having a current flowing to the right, as shown in Figure 20.6. In which direction is the magnetic force exerted on the particle? A) downward B) upward C) into the page D) out of the page E) The magnetic force is zero since the velocity is parallel to the current.
upward Solution: First find B on the wire using the grip right hand rule: B is going into the page. Then, arm points right for velocity, fingers point into the page, thumb naturally points downwards. However, since the particle is negative, the force is in the opposite direction: upwards.
Consider a solenoid of length L, N windings, and radius b (L is much longer than b). A current I is flowing through the wire. If the radius of the solenoid were doubled (becoming 2b), and all other quantities remained the same, the magnetic field: A) would become one half as strong. B) would become twice as strong. C) would remain the same.
would remain the same