Studyledge

Magnetic Fields: Question 6

Syllabus 20.3

Multiple choice A2 1 mark

A proton (charge +e=1.60×10−19 C+e=1.60\times10^{-19}\text{ C}) travels at a constant speed of v=1.0×106 m s−1v=1.0\times10^{6}\text{ m s}^{-1} vertically downward through a region of uniform magnetic field. The field has a flux density of B=0.60 TB=0.60\text{ T} and is directed out of the page, toward the reader.

What is the direction of the magnetic force on the proton at this instant?

Choose an answer to check it, then compare with the worked solution below.
Show worked solution Hide worked solution

Worked solution

Step 1: Set up Fleming’s left-hand rule for a positive charge

For a moving charge, Fleming’s left-hand rule applies directly to a positive charge using its actual velocity direction as the “current” direction (for a negative charge the direction would need reversing first, but the proton’s charge is already positive, so no reversal is needed here):

  • First finger → Field
  • Second finger → velocity of the positive charge
  • Thumb → resulting force

Step 2: Apply it to this proton

The field points out of the page, so the first finger points toward you, out of the page. The proton moves vertically downward, so the second finger points down.

With the first finger out of the page and the second finger pointing down, the thumb, held perpendicular to both, points horizontally to the left, in the plane of the page.

Step 3: Check with the vector form

Using F⃗=qv⃗×B⃗\vec{F}=q\vec{v}\times\vec{B} with axes x^\hat{x} (right), y^\hat{y} (up the page) and z^\hat{z} (out of the page): v⃗=−vy^\vec{v}=-v\hat{y} (downward) and B⃗=Bz^\vec{B}=B\hat{z} (out of page), so F⃗=q(−vy^)×(Bz^)=−qvB(y^×z^)=−qvBx^\vec{F}=q(-v\hat{y})\times(B\hat{z})=-qvB(\hat{y}\times\hat{z})=-qvB\hat{x} Since y^×z^=x^\hat{y}\times\hat{z}=\hat{x}, this gives F⃗=−qvBx^\vec{F}=-qvB\hat{x}, i.e. a force in the −x^-\hat{x} direction, to the left, confirming Step 2.

Step 4: Why the other options are wrong

  • B (to the right): this is the reverse of the correct thumb direction; it would result from mixing up which finger represents the field and which represents the velocity, or from mistakenly reversing the direction because the charge is (wrongly assumed to be) negative.
  • C (out of the page): the force must be perpendicular to both the velocity (vertical, in the plane of the page) and the field (out of the page). The only directions perpendicular to both of these are horizontally left or right, in the plane of the page, not out of the page.
  • D (into the page): for the same reason as C, the force cannot be parallel to the field itself.

Step 5: Checking the magnitude (for completeness)

The velocity is perpendicular to the field, so θ=90°\theta=90° and sin⁡θ=1\sin\theta=1. Using F=BQvsin⁡θF=BQv\sin\theta: F=0.60×1.60×10−19×1.0×106×sin⁡90°=9.6×10−14 NF=0.60\times1.60\times10^{-19}\times1.0\times10^{6}\times\sin90°=9.6\times10^{-14}\text{ N}

Recomputing as a check: 0.60×1.60=0.960.60\times1.60=0.96, and the powers of ten give 10−19+6=10−1310^{-19+6}=10^{-13}, so F=0.96×10−13=9.6×10−14 NF=0.96\times10^{-13}=9.6\times10^{-14}\text{ N}, the same result.

Final answer

  • The magnetic force on the proton is horizontally to the left\boxed{\text{horizontally to the left}}, in the plane of the page (magnitude 9.6×10−14 N9.6\times10^{-14}\text{ N}), option A.