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20.2. Force on a current-carrying conductor
A subsection of Physics, 9702, through 20. Magnetic fields
Listing 10 of 12 questions
A rigid copper wire is held horizontally between the pole pieces of two magnets, as shown in . 8.5 cm 8.5 cm S S N N copper wire direction of current side view top view θ The width of each pole piece is 8.5 cm. The uniform magnetic flux density B in the region between the poles of the magnets is 3.7 mT and is zero outside this region. The angle between the wire and the direction of the magnetic field is θ. The current in the wire is in the direction shown on . By reference to the side view of , state and explain the direction of the force on the magnets. The constant current in the wire is 5.1 A. For angle θ equal to 90°, calculate the force on the wire. force = N The angle θ is changed to 60°. The length of wire in the magnetic field is ° . sin60 8 5 c m cm. Calculate the force on the wire. force = N The constant current in the wire is now changed to an alternating current of frequency 20 Hz and root-mean-square (r.m.s.) value 5.1 A. The angle between the wire and the direction of the magnetic field is 90°. On , sketch a graph to show the variation with time t of the force F on the wire for two cycles of the alternating current. F / N t / s
9702_s18_qp_41
THEORY
2018
Paper 4, Variant 1
View
A rigid copper wire is held horizontally between the pole pieces of two magnets, as shown in . 8.5 cm 8.5 cm S S N N copper wire direction of current side view top view θ The width of each pole piece is 8.5 cm. The uniform magnetic flux density B in the region between the poles of the magnets is 3.7 mT and is zero outside this region. The angle between the wire and the direction of the magnetic field is θ. The current in the wire is in the direction shown on . By reference to the side view of , state and explain the direction of the force on the magnets. The constant current in the wire is 5.1 A. For angle θ equal to 90°, calculate the force on the wire. force = N The angle θ is changed to 60°. The length of wire in the magnetic field is ° . sin60 8 5 c m cm. Calculate the force on the wire. force = N The constant current in the wire is now changed to an alternating current of frequency 20 Hz and root-mean-square (r.m.s.) value 5.1 A. The angle between the wire and the direction of the magnetic field is 90°. On , sketch a graph to show the variation with time t of the force F on the wire for two cycles of the alternating current. F / N t / s
9702_s18_qp_43
THEORY
2018
Paper 4, Variant 3
View
Define the tesla. A magnet produces a uniform magnetic field of flux density B in the space between its poles. A rigid copper wire carrying a current is balanced on a pivot. Part PQLM of the wire is between the poles of the magnet, as illustrated in . pivot magnet N S P L M Q rigid copper wire weight W 5.6 cm (not to scale) The wire is balanced horizontally by means of a small weight W. The section of the wire between the poles of the magnet is shown in . pole of magnet N rigid copper wire Q P L M S pole of magnet (not to scale) Explain why: section QL of the wire gives rise to a moment about the pivot sections PQ and LM of the wire do not affect the equilibrium of the wire. Section QL of the wire has length 0.85 cm. The perpendicular distance of QL from the pivot is 5.6 cm. When the current in the wire is changed by 1.2 A, W is moved a distance of 2.6 cm along the wire in order to restore equilibrium. The mass of W is 1.3 × 10–4 kg. Show that the change in moment of W about the pivot is 3.3 × 10–5 N m. Use the information in to determine the magnetic flux density B between the poles of the magnet. B = T
9702_s20_qp_41
THEORY
2020
Paper 4, Variant 1
View
Define the tesla. A magnet produces a uniform magnetic field of flux density B in the space between its poles. A rigid copper wire carrying a current is balanced on a pivot. Part PQLM of the wire is between the poles of the magnet, as illustrated in . pivot magnet N S P L M Q rigid copper wire weight W 5.6 cm (not to scale) The wire is balanced horizontally by means of a small weight W. The section of the wire between the poles of the magnet is shown in . pole of magnet N rigid copper wire Q P L M S pole of magnet (not to scale) Explain why: section QL of the wire gives rise to a moment about the pivot sections PQ and LM of the wire do not affect the equilibrium of the wire. Section QL of the wire has length 0.85 cm. The perpendicular distance of QL from the pivot is 5.6 cm. When the current in the wire is changed by 1.2 A, W is moved a distance of 2.6 cm along the wire in order to restore equilibrium. The mass of W is 1.3 × 10–4 kg. Show that the change in moment of W about the pivot is 3.3 × 10–5 N m. Use the information in to determine the magnetic flux density B between the poles of the magnet. B = T
9702_s20_qp_43
THEORY
2020
Paper 4, Variant 3
View
State what is meant by a magnetic field. A long, straight wire P carries a current into the page, as shown in . wire P current into page On , draw four field lines to represent the magnetic field around wire P due to the current in the wire. A second long, straight wire Q, carrying a current of 5.0 A out of the page, is placed parallel to wire P, as shown in . wire P current into page wire Q current 5.0 A out of page The flux density of the magnetic field at wire Q due to the current in wire P is 2.6 mT. Calculate the magnetic force per unit length exerted on wire Q by wire P. force per unit length = N m–1 State the direction of the force exerted on wire Q by wire P. The flux density of the magnetic field at wire P due to the current in wire Q is 1.5 mT. Determine the magnitude of the current in wire P. Explain your reasoning. current = A
9702_s23_qp_41
THEORY
2023
Paper 4, Variant 1
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State what is meant by a magnetic field. A long, straight wire P carries a current into the page, as shown in . wire P current into page On , draw four field lines to represent the magnetic field around wire P due to the current in the wire. A second long, straight wire Q, carrying a current of 5.0 A out of the page, is placed parallel to wire P, as shown in . wire P current into page wire Q current 5.0 A out of page The flux density of the magnetic field at wire Q due to the current in wire P is 2.6 mT. Calculate the magnetic force per unit length exerted on wire Q by wire P. force per unit length = N m–1 State the direction of the force exerted on wire Q by wire P. The flux density of the magnetic field at wire P due to the current in wire Q is 1.5 mT. Determine the magnitude of the current in wire P. Explain your reasoning. current = A
9702_s23_qp_43
THEORY
2023
Paper 4, Variant 3
View
A stiff straight copper wire XY is held fixed in a uniform magnetic field of flux density 2.6 × 10−3 T, as shown in . 4.7 cm 34° stiff wire uniform magnetic field flux density 2.6 × 10–3 T X Y 34° The wire XY has length 4.7 cm and makes an angle of 34° with the magnetic field. Calculate the force on the wire due to a constant current of 5.4 A in the wire. force = N The current in the wire is now changed to an alternating current of r.m.s. value 1.7 A. Determine the total variation in the force on the wire due to the alternating current. variation in force = N
9702_w14_qp_41
THEORY
2014
Paper 4, Variant 1
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A stiff straight copper wire XY is held fixed in a uniform magnetic field of flux density 2.6 × 10−3 T, as shown in . 4.7 cm 34° stiff wire uniform magnetic field flux density 2.6 × 10–3 T X Y 34° The wire XY has length 4.7 cm and makes an angle of 34° with the magnetic field. Calculate the force on the wire due to a constant current of 5.4 A in the wire. force = N The current in the wire is now changed to an alternating current of r.m.s. value 1.7 A. Determine the total variation in the force on the wire due to the alternating current. variation in force = N
9702_w14_qp_42
THEORY
2014
Paper 4, Variant 2
View
Define magnetic flux density. A stiff copper wire is balanced horizontally on a pivot, as shown in . 7.5 cm pivot stiff wire P Q S R Sections PQ, QR and RS of the wire are situated in a uniform magnetic field of flux density B produced between the poles of a permanent magnet. The perpendicular distance of PQRS from the pivot is 7.5 cm. When a current of 2.7 A is passed through the wire, a small mass of 45 mg is placed a distance 8.8 cm from the pivot in order to restore the balance of the wire, as shown in . 7.5 cm 8.8 cm 2.7 A 2.7 A pivot stiff wire P Q S R pole pieces of magnet small mass Explain why, when the current is switched on, the current in the sections PQ and RS of the wire does not affect the balance of the wire. The length of section QR of the wire is 1.2 cm. Calculate the magnetic flux density B. B = T
9702_w18_qp_42
THEORY
2018
Paper 4, Variant 2
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Questions Discovered
12