Figure 2 shows two magnets, supported on a yoke, placed on an electronic balance - AQA - A-Level Physics - Question 2 - 2017 - Paper 2
Question 2
Figure 2 shows two magnets, supported on a yoke, placed on an electronic balance.
Figure 3 shows a simplified plan view of the copper wire and magnets.
The magnets... show full transcript
Worked Solution & Example Answer:Figure 2 shows two magnets, supported on a yoke, placed on an electronic balance - AQA - A-Level Physics - Question 2 - 2017 - Paper 2
Step 1
Which of the following correctly describes the direction of the force acting on the wire DE due to the magnetic field when the switch is closed?
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The correct answer is: vertically up. According to Fleming's Left Hand Rule, the direction of the force is perpendicular to both the current and the magnetic field.
Step 2
Label the poles of the magnets by putting N or S on each of the two dashed lines in Figure 3.
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The left magnet should be labeled 'N' (North) and the right magnet 'S' (South). This configuration indicates the direction of the magnetic field from North to South.
Step 3
Define the tesla.
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The tesla (T) is the SI unit of magnetic flux density. It is defined as the amount of magnetic flux (measured in webers) passing through a unit area (measured in square meters) perpendicular to the direction of the magnetic field. One tesla is equivalent to one weber per square meter.
Step 4
Calculate the magnetic flux density between the magnets.
98%
120 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Using the formula for magnetic force: B=I⋅lF, where ( F = mg ) with ( m = 0.620 g = 0.000620 kg ), ( g = 9.81 ,m/s^2 ), ( I = 3.43 , A ), and ( l = 0.0500 , m ):
Calculate the force:
F=0.000620kg⋅9.81m/s2=0.0060782N
Substitute into the formula for B:
B=3.43A⋅0.0500m0.0060782N=0.0350T≈0.036T