The Physics Question Bank

Course: Combined Science + Separate Physics  |  Tier: Higher only  |  Equation sheet: given in the exam

The equation

force on a conductor carrying a current = magnetic flux density × current × length (at right angles to the field)    F = B I l

Quantity	Symbol	Unit
force	F	newtons (N)
magnetic flux density	B	tesla (T)
current	I	amperes (A)
length of conductor in the field	l	metres (m)

Rearranged: B = F ÷ (I l)  |  I = F ÷ (B l)  |  l = F ÷ (B I)  |  1 mT = 0.001 T

Use FIFA for every answer

• F – Formula: write the equation as it appears on the sheet.
• I – Insert: put in the numbers (convert to standard units first).
• F – Fix: rearrange to make the unknown the subject.
• A – Answer: calculate and always give the unit.

Higher tier only: work through all 10. Most questions need rearranging.

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Questions

Q1 A 0.5 m wire carries 3 A at right angles to a 0.2 T magnetic field. Calculate the force on the wire.

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F: F = B I l
I: F = 0.2 × 3 × 0.5
F: F is already the subject
A: 0.3 N

Q2 A 0.5 m wire in a 0.4 T field feels a force of 0.6 N. Calculate the current.

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F: F = B I l
I: 0.6 = 0.4 × I × 0.5, so 0.6 = 0.2 × I
F: I = 0.6 ÷ 0.2
A: 3 A

Q3 A 0.3 m wire carrying 2 A feels a force of 0.12 N. Calculate the magnetic flux density.

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F: F = B I l
I: 0.12 = B × 2 × 0.3, so 0.12 = 0.6 × B
F: B = 0.12 ÷ 0.6
A: 0.2 T

Q4 A wire carrying 4 A in a 0.15 T field feels a force of 0.12 N. What length of wire is in the field, in cm?

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F: F = B I l
I: 0.12 = 0.15 × 4 × l, so 0.12 = 0.6 × l
F: l = 0.12 ÷ 0.6 = 0.2 m
A: 20 cm

Q5 A wire carrying 2.5 A in a 50 mT field feels a force of 0.025 N. Calculate the length of wire in the field.

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Convert first: 50 mT = 0.05 T
F: F = B I l
I: 0.025 = 0.05 × 2.5 × l, so 0.025 = 0.125 × l
F: l = 0.025 ÷ 0.125
A: 0.2 m

Q6 A loudspeaker coil has 5.0 m of wire in a 0.8 T field. It needs a force of 2.0 N. Calculate the current needed.

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F: F = B I l
I: 2.0 = 0.8 × I × 5.0, so 2.0 = 4.0 × I
F: I = 2.0 ÷ 4.0
A: 0.5 A

Q7 10 cm of wire carrying 800 mA feels a force of 0.024 N. Calculate the magnetic flux density.

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Convert first: 10 cm = 0.1 m; 800 mA = 0.8 A
F: F = B I l
I: 0.024 = B × 0.8 × 0.1, so 0.024 = 0.08 × B
F: B = 0.024 ÷ 0.08
A: 0.3 T

Q8 A power line carries 1.0 × 10³ A across the Earth’s magnetic field (5.0 × 10⁻⁵ T). The force on a section of the line is 10 N. How long is the section?

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F: F = B I l
I: 10 = 5.0 × 10⁻⁵ × 1.0 × 10³ × l, so 10 = 0.05 × l
F: l = 10 ÷ 0.05
A: 2.0 × 10² m

Q9 One side of a motor coil is 0.05 m long and the coil has 100 turns. It carries 1.5 A in a 0.2 T field. Calculate the total force on that side of the coil.

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Total length in the field: 100 × 0.05 = 5 m
F: F = B I l
I: F = 0.2 × 1.5 × 5
F: F is already the subject
A: 1.5 N

Q10 0.12 m of wire in a 0.35 T field feels a force of 0.090 N. Calculate the current, to 2 significant figures.

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F: F = B I l
I: 0.090 = 0.35 × I × 0.12, so 0.090 = 0.042 × I
F: I = 0.090 ÷ 0.042 = 2.14… A
A: 2.1 A (2 s.f.)