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Questions on magnetic flux density and the force on a wire
The motor effect is the sideways force on a current-carrying wire in a magnetic field, F = BIl for a wire at right angles to the field. Fleming's left-hand rule gives the direction, and rearranging to B = F/Il defines magnetic flux density and the tesla. In the required practical, a top-pan balance measures the force.
18 original questions · 47 marks · the magnetic flux density and the force on a wire notes · Magnetic fields
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State the equation for the force on a current-carrying wire in a magnetic field and the condition under which it applies.
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F = BIl, where B is the flux density, I the current and l the length of wire in the field (1); valid when the wire is perpendicular to the field (1).A wire of length 0.15 m carries a current of 3.0 A and lies perpendicular to a magnetic field of flux density 0.20 T. Calculate the force on the wire.
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F = BIl = 0.20 × 3.0 × 0.15 (1)
F = 0.090 N (1)Define magnetic flux density and express the tesla in terms of other SI units.
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The force per unit current per unit length on a wire carrying a current at right angles to the field, B = F/Il (1); 1 T = 1 N A−1 m−1 (1).On diagrams of magnetic fields, state what is represented by a circled cross and by a circled dot.
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A circled cross represents a field directed into the page and a circled dot a field directed out of the page (1).A wire carrying a current of 2.5 A passes between the poles of a U-shaped magnet. A 45 mm length of the wire lies at right angles to the field, which has flux density 90 mT. Calculate the force on the wire.
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F = BIl = 0.090 × 2.5 × 0.045 (1)
F = 0.010 N (1)A 0.25 m length of wire carrying a current of 4.0 A experiences a force of 0.50 N when placed perpendicular to a magnetic field. Calculate the magnetic flux density.
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B = F/(Il) = 0.50/(4.0 × 0.25) (1)
B = 0.50 T (1)A wire of length 0.10 m carrying a current of 2.0 A is placed at 30° to a magnetic field of flux density 0.30 T. Calculate the force on the wire.
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F = BIl sinθ = 0.30 × 2.0 × 0.10 × sin 30° (1)
F = 0.030 N (1)A wire of length 0.080 m lies perpendicular to a magnetic field of flux density 0.45 T and experiences a force of 0.18 N. Calculate the current in the wire.
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I = F/(Bl) = 0.18/(0.45 × 0.080) (1)
I = 5.0 A (1)Explain why the force on a current-carrying wire in a magnetic field is a maximum when the wire is perpendicular to the field and zero when it is parallel to the field.
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The force depends on the angle θ between the wire and the field through F = BIl sinθ (1). At 90°, sinθ = 1, giving the maximum force (1); when the wire is parallel to the field, θ = 0 and sinθ = 0, so there is no force (1).Two long, straight, parallel wires carry currents in opposite directions. State whether the wires attract or repel each other, and explain your answer by considering the force on one wire in the magnetic field of the other.
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They repel (1). Each wire lies in the circular field of the other and so feels a force F = BIl; applying Fleming's left hand rule to either wire, using the field of its neighbour, gives a force directed away from the neighbour (1).A rigid wire of length 0.060 m and mass 0.40 g rests on two supports between the poles of a magnet. The horizontal magnetic field of flux density 0.15 T is perpendicular to the wire. Calculate the current for which the magnetic force just supports the weight of the wire.
g = 9.81 m s−2Mark scheme
Weight = mg = 0.40 × 10−3 × 9.81 = 3.92 × 10−3 N (1)
For support, BIl = mg, so I = mg/(Bl) = 3.92 × 10−3/(0.15 × 0.060) (1)
I = 0.44 A (1)A straight wire of length 70 mm carries a current of 3.2 A at 25° to a uniform magnetic field. The force on the wire is 8.4 × 10−3 N. Calculate the flux density of the field.
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B = F/(Il sinθ) (1)
= (8.4 × 10−3)/(3.2 × 0.070 × sin 25°) (1)
B = 0.089 T (1)In a current balance, a horizontal wire of length 0.050 m carrying a current of 8.0 A lies perpendicular to a magnetic field of flux density 0.60 T, so that the force on it is vertical. Calculate the force on the wire and the mass that has the same weight.
g = 9.81 m s−2Mark scheme
F = BIl = 0.60 × 8.0 × 0.050 (1)
F = 0.24 N (1)
m = F/g = 0.24/9.81 (1)
m = 24.5 g (1)A wire of length 0.12 m carries a current of 6.0 A in a magnetic field of flux density 0.25 T. Calculate the force on the wire when it is perpendicular to the field, and when it is at 45° to the field.
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Perpendicular: F = BIl = 0.25 × 6.0 × 0.12 (1)
F = 0.18 N (1)
At 45°: F = BIl sin 45° = 0.18 × sin 45° (1)
F = 0.127 N (1)Describe how the force on a current-carrying wire can be measured in the laboratory using a top-pan balance.
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Clamp a horizontal wire between the poles of a magnet standing on a top-pan balance, with the wire perpendicular to the field; zero the balance, then pass a known current (1). The wire exerts an equal and opposite force on the magnet (Newton's third law), changing the balance reading (1). Force = change in mass reading (in kg) × g; repeat for a range of currents (1).In the arrangement of required practical 10, a student measures the force on a wire for a range of currents, keeping 120 mm of the wire in the field throughout. The graph of force against current is a straight line through the origin with gradient 0.048 N A−1. Explain why the gradient equals Bl and determine the flux density of the field.
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F = BIl, so with B and l constant the graph of F against I is a straight line through the origin with gradient Bl (1)
B = gradient/l = 0.048/0.120 (1)
B = 0.40 T (1)A design for an actuator requires a force of at least 0.15 N on a straight wire held perpendicular to a magnetic field of flux density 0.28 T. Four wires are available.
Wire W: length in field 45 mm, maximum working current 8.0 A
Wire X: length in field 60 mm, maximum working current 10.0 A
Wire Y: length in field 85 mm, maximum working current 5.0 A
Wire Z: length in field 70 mm, maximum working current 9.0 A
Deduce which of the wires could provide the required force at their maximum working current.Mark scheme
Use of F = BIl with each length in metres at the maximum current (1)
W: F = 0.101 N and Y: F = 0.119 N (1)
X: F = 0.168 N and Z: F = 0.176 N (1)
Only X and Z reach 0.15 N, so either is suitable; W and Y are not (1)A magnet stands on a top-pan balance and a fixed horizontal wire passes between its poles at right angles to the field, with 50 mm of the wire in the field. When the current in the wire is 3.5 A, the balance reading increases by 0.65 g. Determine the flux density of the field.
g = 9.81 m s−2Mark scheme
By Newton's third law the force on the magnet equals the force on the wire: F = Δm g = 0.65 × 10−3 × 9.81 = 6.4 × 10−3 N (1)
B = F/(Il) = 6.4 × 10−3/(3.5 × 0.050) (1)
B = 0.036 T (1)
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