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Omniguides by NOX
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Chapter 5 Solutions

These worked solutions are authored for the reviewed prompts shown here. Any added geometry or assumption is stated. Original chapter problems supplied no answer key, so these solutions are separate from the archived manuscript.

Original problem 1 · worked solution

A velocity selector has E⃗=Ey^\vec E=E\hat y, B⃗=Bz^\vec B=B\hat z and incoming velocity v⃗=vx^\vec v=v\hat x, with E,B>0. Find the speed for an undeflected charged particle.

Authored clarification: The original missing “blah” diagram is replaced by explicit crossed-field directions. Assume a nonzero point charge and negligible gravity/collisions.

  1. Add electric and magnetic forces.

    F⃗=q(E⃗+v⃗×B⃗)=q(E−vB)y^\vec F=q(\vec E+\vec v\times\vec B)=q(E-vB)\hat y.

  2. Require force cancellation.

    E−vB=0⇒v=E/BE-vB=0\quad\Rightarrow\quad v=E/B.
Answer

The selected speed is E/B for either charge sign with the stated direction of motion.

Check

E/B has units m/s. Reversing the velocity without changing the fields would make the two forces add.

Practice prompt · Original reading location

Archived original prompt
  • A velocity selector is a device which uses both magnetic and electric fields in order to pick out charged particle which have a particular velocity as pictured in blah. If a charged particle of charge q is shot through such a device with an electric field E and magnetic field B what must the velocity of the charged particle be in order for a particle to pass through the velocity selector while moving in a straight line?

    • Original problem 2 · worked solution

    A nonrelativistic ion of mass m and charge magnitude |q| enters a uniform field B perpendicular to its velocity v. It follows a semicircle and strikes a detector in the entrance plane. Find the distance between entrance and impact points.

    Authored clarification: The original detector diagram is missing. A semicircular path and detector placement are explicit authored conditions.

    1. Magnetic force supplies the centripetal force.

      ∣q∣vB=mv2/r|q|vB=mv^2/r.

    2. Solve the orbital radius.

      r=mv∣q∣Br={mv\over|q|B}.

    3. A semicircle returns to the entrance plane at the other end of its diameter.

      D=2r=2mv∣q∣BD=2r={2mv\over|q|B}.
    Answer

    D=2mv/(∣q∣B)D=2mv/(|q|B). The charge sign determines which side of the entrance the ion hits.

    Check

    The result is conditional on a half-circle detector geometry; another detector placement would give another displacement.

    Practice prompt · Original reading location

    Archived original prompt
  • A mass spectrometer is a device which sorts charged particles based on their masses. To do this particles of a substance are ionized then accelerated to a certain velocity (a velocity selector ensures they have a certain velocity) and then shot into a region with uniform magnetic field. Using diagram blah determine how far away a particle of velocity v with mass m and charge q in a uniform magnetic field B will hit away from the starting point of the particle.

    • Original problem 3 · worked solution

    Two uranium isotope ions have molar masses 235 g/mol and 238 g/mol and the same nonzero signed charge q. From a shared entry point they enter the same uniform magnetic field of magnitude B>0 with identical parallel incident velocity of speed v>0, perpendicular to the field. Their speeds are nonrelativistic. Each follows a semicircle back to the common entrance-plane detector, so both bend to the same side. Find their impact separation.

    Authored clarification: The source lacks speed, field, signed charge state, common entry, beam orientation and detector geometry. This authored model states all of those symbolically, with the same charge sign and nonrelativistic speeds; no numerical separation is inferred.

    1. Convert molar-mass difference to one-ion mass difference.

      Δm=0.003 kg/molNA≈4.98×10−27 kg\Delta m={0.003\,\mathrm{kg/mol}\over N_A}\approx4.98\times10^{-27}\,\mathrm{kg}.

    2. Use each semicircle’s diameter for the stated nonrelativistic, common-entry, same-sign beam.

      Di=2miv∣q∣BD_i={2m_iv\over|q|B}.

      Both ions have the same incident direction and the same signed charge, so they bend to the same side of the shared entry point.

    3. Subtract the two same-side impact positions.

      ΔD=2vΔm∣q∣B\Delta D={2v\Delta m\over|q|B}.
    Answer

    ΔD=2v(0.003 kg/mol)NA∣q∣B\Delta D={2v(0.003\,\mathrm{kg/mol})\over N_A|q|B}. For singly charged ions, use |q|=e.

    Check

    The heavier isotope has the larger radius at equal nonrelativistic speed and identical signed charge. The difference formula assumes the shared entry and parallel beam stated above. Opposite charge signs would place impacts on opposite sides and give the sum of diameters; different entry points would add their displacement. Without v and B, a numerical separation is unavailable.

    Practice prompt · Original reading location

    Archived original prompt
  • The first atomic weapon used in World War 2 required extremely pure U-235, an isotope of uranium. Since a sample of uranium contains several different isotopes mass spectrometers were used in order to retrieve pure U-235 due to the different isotopes small differences in mass. If the mass of U-238, the main isotope in a sample of uranium, is 238 g/mole and the mass of U-235 is 235 g/mole how far apart will the two isotopes land in a mass spectrometer?

    • Original problem 4 · worked solution

    A 30 kg child carries 2 fC of net charge in a 1 T uniform magnetic field. Even with the velocity perpendicular to B and force directed upward, can magnetic force balance the child’s weight?

    Authored clarification: Model the net charge as a point charge, use g=9.8 m/s², and take the velocity perpendicular to B with the magnetic force directed upward.

    1. The largest force for a given speed is |q|vB.

      mg=30(9.8)=294 Nmg=30(9.8)=294\,\mathrm N.

    2. Compute the formal required speed.

      v=mg∣q∣B=2942×10−15=1.47×1017 m/sv={mg\over|q|B}={294\over2\times10^{-15}}=1.47\times10^{17}\,\mathrm{m/s}.

    3. Compare to a physical bound.

      v/c≈4.90×108,∣q∣cB≈6.00×10−7 N≪294 Nv/c\approx4.90\times10^8,\qquad |q|cB\approx6.00\times10^{-7}\,\mathrm N\ll294\,\mathrm N.
    Answer

    No attainable speed balances the weight with these parameters. The formal classical value exceeds c by about 490 million times.

    Check

    Changing the motion direction can only reduce the magnetic-force magnitude below the perpendicular value.

    Source wording and interpretation: The formal nonrelativistic force balance must be checked against the speed of light; it is not a feasible levitation scenario.

    Practice prompt · Original reading location

    Archived original prompt
  • A small child weighing around 30 kg gathers 2 fC of charge on his socks. If the child is running inside a machine which produces a 1 T magnetic field how fast must the child be running to lift off from the ground assuming gravity to be 9.8 m/s²?

    • Original problem 5 · worked solution

    A closed wire loop carries steady current I in a spatially uniform magnetic field B. Determine the net magnetic force, and state whether zero force guarantees no motion.

    Authored clarification: The field is uniform over the whole loop. The question asks net force, not torque or force on each segment.

    1. Integrate the force on current elements.

      F⃗=I∮dℓ⃗×B⃗\vec F=I\oint d\vec\ell\times\vec B.

    2. Factor out the uniform field.

      F⃗=I(∮dℓ⃗)×B⃗=0\vec F=I\left(\oint d\vec\ell\right)\times\vec B=0.

    3. Distinguish torque and deformation from net force.

      A loop can still experience torque τ⃗=μ⃗×B⃗\vec\tau=\vec\mu\times\vec B or internal stresses. Zero net force alone does not prohibit rotation or deformation.
    Answer

    The net magnetic force is zero. Motion requires analysis of torque, rigidity, constraints and initial conditions.

    Check

    The conclusion is specific to a uniform field; nonuniform fields may exert a net force.

    Practice prompt · Original reading location

    Archived original prompt
  • Consider any closed loop of wire with current I in a uniform magnetic field. Determine the force acting on the wire.

    • Original problem 6 · worked solution

    A horizontal wire carries 15 A in a uniform horizontal magnetic field of magnitude 0.5 μT. Its current is perpendicular to B and oriented so the magnetic force is upward. What mass per unit length can this force support for g=9.8 m/s²?

    Authored clarification: The source does not fully specify orientation. For example, take current east and field north; east×north is upward.

    1. Magnetic force per unit length is IB for perpendicular directions.

      F/ℓ=(15)(0.5×10−6)=7.5×10−6 N/mF/\ell=(15)(0.5\times10^{-6})=7.5\times10^{-6}\,\mathrm{N/m}.

    2. Balance the weight per unit length.

      λmg=IB⇒λm=IBg=7.65×10−7 kg/m\lambda_mg=IB\quad\Rightarrow\quad\lambda_m={IB\over g}=7.65\times10^{-7}\,\mathrm{kg/m}.
    Answer

    The supported mass per unit length is 7.65×10⁻⁷ kg/m, about 0.765 mg/m, with upward Lorentz force.

    Check

    The tiny supported mass reflects the very weak magnetic field.

    Practice prompt · Original reading location

    Archived original prompt
  • A straight wire with a current of 15 amps is placed in a magnetic field of 0.5 μT. If we want this wire to levitate near the earths surface what how must the magnetic field and wire be pointing relative to the surface and what must the mass per length of the wire be?

    • Original problem 7 · worked solution

    Optional original-source extension. Prerequisites: a vector line integral and the endpoint-displacement identity in a uniform field. This item makes no AP coverage claim.

    A wire follows y=Lex/Ly=L e^{x/L} in the xy plane from x=0 to x=a, with L>0 and current I toward increasing x. A uniform field B⃗=Bz^\vec B=B\hat z is applied. Find the net magnetic force and its magnitude.

    Authored clarification: The source y=eˣ needs a length scale. Here L gives the exponential a dimensionless argument; it is an explicit authored parameter rather than an inferred source value.

    1. For a uniform field, only endpoint displacement enters the net force.

      F⃗=I∫dℓ⃗×B⃗=I(r⃗f−r⃗i)×B⃗\vec F=I\int d\vec\ell\times\vec B=I(\vec r_f-\vec r_i)\times\vec B.

    2. Find the endpoint displacement.

      Δr⃗=ax^+L(ea/L−1)y^\Delta\vec r=a\hat x+L(e^{a/L}-1)\hat y.

    3. Compute the cross product and norm.

      F⃗=IB[L(ea/L−1)x^−ay^],F=∣IB∣a2+L2(ea/L−1)2\vec F=IB[L(e^{a/L}-1)\hat x-a\hat y],\quad F=|IB|\sqrt{a^2+L^2(e^{a/L}-1)^2}.
    Answer

    The displayed vector and magnitude are the net force; they depend on the chord, not the wire’s arc length.

    Check

    For a=0 the displacement and net force vanish. Each term under the square root has units m².

    Practice prompt · Original reading location

    Archived original prompt
  • A wires shape is determine by the curve of eˣ on a domain of [0, x]. This piece of wire is laid down on a desk and a uniform magnetic field is produced which points upwards and perpendicular to the desk with a magnitude B. If the current running through the wire is given by I what is the magnitude of the force acting on the wire?
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