Photo AI

15 A student placed two magnets on a holder so that the north pole of one magnet faced the south pole of the other magnet - Edexcel - A-Level Physics - Question 15 - 2023 - Paper 4

Question icon

Question 15

15-A-student-placed-two-magnets-on-a-holder-so-that-the-north-pole-of-one-magnet-faced-the-south-pole-of-the-other-magnet-Edexcel-A-Level Physics-Question 15-2023-Paper 4.png

15 A student placed two magnets on a holder so that the north pole of one magnet faced the south pole of the other magnet. The arrangement was placed on a sensitive ... show full transcript

Worked Solution & Example Answer:15 A student placed two magnets on a holder so that the north pole of one magnet faced the south pole of the other magnet - Edexcel - A-Level Physics - Question 15 - 2023 - Paper 4

Step 1

When the student moved the wire quickly downwards between the magnets, how the reading on the balance changed

96%

114 rated

Answer

When the student moved the wire downward rapidly within the magnetic field created by the magnets, electromagnetic induction occurred due to the change in magnetic flux through the loop of wire. According to Faraday's Law of Electromagnetic Induction, a change in magnetic flux induces an electromotive force (emf) in the wire. This induced emf interacts with the magnetic field, generating a force on the wire as per the Lorentz force principle.

As the wire quickly moves through the magnetic field, it experiences a force that acts against the direction of its motion, effectively increasing the force acting on the balance. This results in an increase in the reading on the balance. If the wire moves slowly or stays still, the magnetic flux does not change significantly, and thus, the force on the balance would remain stable.

Step 2

Calculate the maximum p.d. that could have been measured

99%

104 rated

Answer

To calculate the maximum potential difference (p.d.), we can use the formula: V=BimesLimesvV = B imes L imes v

Where:

  • VV = potential difference (p.d.)
  • BB = magnetic flux density (0.35 T)
  • LL = length of the wire within the magnetic field (34 mm or 0.034 m)
  • vv = vertical speed of the wire (2.2 m/s)

Substituting the values: V=0.35imes0.034imes2.2V = 0.35 imes 0.034 imes 2.2

Calculating this:

  1. Calculate 0.35imes0.034=0.01190.35 imes 0.034 = 0.0119.
  2. Now, multiply by 2.22.2: 0.0119imes2.2=0.026180.0119 imes 2.2 = 0.02618 V.

Thus, the maximum potential difference (p.d.) that could have been measured is approximately:

Vext(maximum)hickapprox0.026extVor26extmVV ext{ (maximum)} hickapprox 0.026 ext{ V or } 26 ext{ mV}

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;