Magnetic Forces and Charged-Particle Motion: questions with solutions
3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
A particle of mass m and charge −q (q>0) is at the origin at t=0 with velocity v0x^ in a uniform magnetic field B=B0z^ (B0>0) that fills all space. The axes x,y,z are right-handed and gravity is neglected. Let R=qB0mv0. The position of the particle at t=3qB0πm is ______. Exactly one option is correct.
A(23R,−2R,0)
B(2R,(1−23)R,0)
C(23R,2R,0)
D(−23R,2R,0)
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Answer: C
Force at t=0: F=(−q)v0x^×B0z^=(−q)v0B0(−y^)=+qv0B0y^. So the particle curves towards +y and the centre of its circle (radius R=mv0/(qB0)) is at (0,R,0).
Starting at the lowest point of this circle moving along +x, it goes anticlockwise as seen from +z (a negative charge in a field along +z circulates anticlockwise).
Angular speed ω=qB0/m, so at the given time it has turned through ωt=π/3. Position: x=Rsinωt=23R, y=R(1−cosωt)=2R, z=0.
Check: the speed stays v0 (the force is perpendicular to the velocity); a positive charge would give (23R,−2R,0), which is the sign trap.
Difficulty: Hard
Question 2Level 3 · AdvancedSingle correct
A uniform magnetic field B=Bz^ (B>0) exists only in the slab 0≤x≤d; there is no field elsewhere. A particle of mass m and charge +q enters the slab at the origin with speed v. Its velocity lies in the xy-plane, makes an angle α (0<α<π/2) with the +x axis and has a positive y-component. Gravity is neglected. The particle crosses the far face x=d if and only if v exceeds ______. Exactly one option is correct.
Am(1+sinα)qBd
Bm(1−sinα)qBd
CmcosαqBd
Dm(1+cosα)qBd
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Answer: A
At entry, v=v(cosα,sinα,0) and F=qv×Bz^=qvB(sinα,−cosα,0). The force points to the centre of the circular path (radius r=mv/(qB)), so the centre is at C=r(sinα,−cosα) and the motion is clockwise seen from +z.
Moving clockwise from the origin, x keeps increasing until the velocity points along −y, i.e. at the rightmost point of the circle, xmax=xC+r=r(1+sinα). After that x decreases and the particle returns to x=0.
So it crosses x=d if and only if r(1+sinα)>d, i.e. v>m(1+sinα)qBd.
Checks: α→0 gives r>d (normal entry). If the velocity had a negative y-component the centre would be closer to the boundary and the factor would be (1−sinα) (option B, the direction trap). Speed stays v throughout since the magnetic force does no work (confirmed by numerical integration of the trajectory).
Difficulty: Peak challenge
Question 3Level 3 · AdvancedSingle correct
A uniform magnetic field of magnitude B, perpendicular to the xy-plane, exists only inside the circular region x2+y2≤R02; outside it the field is zero. A particle of mass m and charge q>0 moving in the xy-plane enters this region with speed v=m3qBR0, its velocity at entry pointing exactly towards the centre of the circle. Gravity is neglected. The time the particle spends inside the field region is ______. Exactly one option is correct.
A6qBπm
B3qB2πm
C2qBπm
D3qBπm
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Answer: D
Inside the region the path is a circular arc of radius r=mv/(qB)=3R0. Let O be the centre of the field region, P the entry point and C the centre of the arc. Since the velocity at P points along PO, the radius CP is perpendicular to OP.
The exit point Q is the second intersection of the two circles; it is the mirror image of P in the line OC, so CQ⊥OQ too: a particle that enters radially leaves radially.
In the right triangle OPC, tan(θ/2)=OP/CP=R0/r=1/3, where θ is the angle turned (equal to the deflection). So θ=60∘=π/3.
Time inside =θ/ω=3π⋅qBm=3qBπm (numerical integration agrees).
Difficulty: Very hard
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