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    • 18.17: Magnetic Forces and Fields (Answers)
      https://phys.libretexts.org/Courses/Kettering_University/Electricity_and_Magnetism_with_Applications_to_Amateur_Radio_and_Wireless_Technology/18%3A_Calculation_of_Magnetic_Quantities_from_Currents/18.17%3A_Magnetic_Forces_and_Fields_(Answers)
      The magnitude of the proton and electron magnetic forces are the same since they have the same amount of charge. 43. A⋅m2⋅T=A⋅m2.NA⋅m=N⋅m Since E=Blv, wh...The magnitude of the proton and electron magnetic forces are the same since they have the same amount of charge. 43. A⋅m2⋅T=A⋅m2.NA⋅m=N⋅m Since E=Blv, where the width is twice the radius, I=2r,I=2r,I=nqAvd,vd=InqA=Inqπr2 so E=B×2r×Inqπr2=2IBnqπr∝1r∝1d. The Hall voltage is inversely proportional to the diameter of the wire.
    • 7.A: Magnetic Forces and Fields (Answers)
      https://phys.libretexts.org/Courses/Muhlenberg_College/Physics_122%3A_General_Physics_II_(Collett)/07%3A_Magnetic_Forces_and_Fields/7.A%3A_Magnetic_Forces_and_Fields_(Answers)
      The magnitude of the proton and electron magnetic forces are the same since they have the same amount of charge. 43. A⋅m2⋅T=A⋅m2.NA⋅m=N⋅m Since E=Blv, wh...The magnitude of the proton and electron magnetic forces are the same since they have the same amount of charge. 43. A⋅m2⋅T=A⋅m2.NA⋅m=N⋅m Since E=Blv, where the width is twice the radius, I=2r,I=2r,I=nqAvd,vd=InqA=Inqπr2 so E=B×2r×Inqπr2=2IBnqπr∝1r∝1d. The Hall voltage is inversely proportional to the diameter of the wire.
    • 8.13: Magnetic Forces and Fields (Answers)
      https://phys.libretexts.org/Courses/Kettering_University/Electricity_and_Magnetism_with_Applications_to_Amateur_Radio_and_Wireless_Technology/08%3A_The_Magnetic_Field/8.13%3A_Magnetic_Forces_and_Fields_(Answers)
      The magnitude of the proton and electron magnetic forces are the same since they have the same amount of charge. 43. A⋅m2⋅T=A⋅m2.NA⋅m=N⋅m Since E=Blv, wh...The magnitude of the proton and electron magnetic forces are the same since they have the same amount of charge. 43. A⋅m2⋅T=A⋅m2.NA⋅m=N⋅m Since E=Blv, where the width is twice the radius, I=2r,I=2r,I=nqAvd,vd=InqA=Inqπr2 so E=B×2r×Inqπr2=2IBnqπr∝1r∝1d. The Hall voltage is inversely proportional to the diameter of the wire.

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