Define the tesla.
- (a)Define the tesla.[3]
- (b)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 Fig. 8.1. [Figure 8.1] 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 Fig. 8.2. [Figure 8.2] Explain why: (i) section QL of the wire gives rise to a moment about the pivot[3]
- (b)(ii)sections PQ and LM of the wire do not affect the equilibrium of the wire.[2]
- (c)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. (i) Show that the change in moment of W about the pivot is 3.3 × 10⁻⁵ Nm.[2]
- (c)(ii)Use the information in (i) to determine the magnetic flux density B between the poles of the magnet.[3]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.
- [B1] magnetic field normal to current
- [B1] newton per ampere
- [B1] newton per metre
- [B1] current in wire QL gives rise to a force / wire QL is perpendicular to the magnetic field
- [B1] force on wire QL is vertical
- [B1] force does not act through the pivot
- [B1] forces act through the same line / forces are horizontal
- [B1] forces are equal (in magnitude) and opposite (in direction)
- [C1] change = mg × (Δ)L
- [A1] = 1.3 x 10⁻⁴ × 9.81 × 2.6 × 10⁻² = 3.3 × 10⁻⁵ N m⁻¹
- [C1] change = B × (Δ)I × L × x
- [C1] 3.3 x 10⁻⁵ = B × 1.2 × 0.85 × 10⁻² × 5.6 × 10⁻²
- [A1] B = 0.058 T
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