This question is about using a digital balance to investigate the force on a wire placed in a magnetic field when there is an electric current in the wire.
A student carries out the following procedure:
- A uniform bar of length 120.0 cm120.0\text{ cm}120.0 cm is pivoted at the 10.0 cm10.0\text{ cm}10.0 cm mark and a triangular support prism is placed on a digital balance.
- One end of the bar is supported by the prism, with the apex of the prism at the 30.0 cm30.0\text{ cm}30.0 cm mark. The height of the pivot is adjusted so that the bar is perfectly horizontal.
- The balance now reads 250.00 g250.00\text{ g}250.00 g.
Deduce the mass of the bar. State one assumption you make.
The student then attaches a uniform wire along the upper edge of the bar. The ends of the wire are connected to terminal blocks on the laboratory bench.
A horizontal uniform magnetic field is applied, perpendicular to the wire, in a localized region between the 87.5 cm87.5\text{ cm}87.5 cm and 92.5 cm92.5\text{ cm}92.5 cm marks on the bar.
The balance reading MMM decreases linearly as the current III in the wire increases:
- At a current of I=1.5 AI = 1.5\text{ A}I=1.5 A, the balance reading is 265.50 g265.50\text{ g}265.50 g.
- At a current of I=5.5 AI = 5.5\text{ A}I=5.5 A, the balance reading is 261.90 g261.90\text{ g}261.90 g.
It can be shown that BBB, the magnitude of the magnetic flux density of the horizontal uniform magnetic field, is given by:
B=σ4LB = \frac{\sigma}{4L}B=4Lσ
where:
- σ\sigmaσ is the magnitude of the change in force acting on the prism per unit current in the wire.
- LLL is the length of the region where the magnetic field cuts through the wire.
Determine BBB. Use g=9.81 m s−2g = 9.81\text{ m s}^{-2}g=9.81 m s−2.