2023 A/L Physics Past Paper · Question 21 (English Medium)

Answer: (5) the proton will continue to move with velocity $v$ along the axis.

What the question asks

We want: what the proton does when it is shot along the axis of a long solenoid that carries a current.

What each letter means:

$q$ ($\text{C}$): the charge of the proton.

$v$ ($\text{m s}^{-1}$): the speed of the proton.

$B$ ($\text{T}$): the magnetic flux density inside the solenoid.

$\theta$ ($^\circ$): the angle between the velocity of the proton and the magnetic field.

$F$ ($\text{N}$): the magnetic force on the proton.

Relevant theory

The magnetic flux density along the axis of a long solenoid carrying a current.

If the moving charge is considered to be equivalent to a current $I$ flowing along a conductor of length $l$ placed along the direction of the motion of the charge, then the force acting on it,

$F = BIl \sin\theta$

$= \dfrac{Bq}{t} Vt \sin\theta$

$F = BqV \sin\theta$

If the charge moves parallel to the field,

$F = 0$

The first quote says the magnetic field of a long solenoid is along its axis. The second quote gives the force on a moving charge. The third quote says the force is zero when the charge moves along the field.

The proton moves along the axis. So it moves along the field.

For example, think of a gardener who pulls a broom along a row of seedlings. A sideways nudge to the seedlings comes from the part of the move that goes across the row.

If the broom goes exactly along the row, there is no across part, so there is no sideways nudge.

The proton is like the broom along the row. It has no part of its motion across the field.

Applying it to this question

$F = BqV\sin\theta$

$\theta = 0^\circ$

$F = BqV\sin 0^\circ = 0$

$0$

$\sin 0^\circ$

$F = BqV \times 0$

Reasoning and calculation

What a zero force means:

With no force the proton gets no acceleration, so its speed does not change. Its direction does not change either.

So the proton keeps moving with the velocity $v$ along the axis.

Check the other end. If the proton moved across the field, with $\theta = 90^\circ$, the force would be $BqV$, the largest it can be. Our angle is $0^\circ$, the other end, where the force is zero.

Reading the options

Test each option against the force.

(1) Speeding up needs a force along the axis. There is none.

(2) Slowing down needs a force along the axis. There is none.

(3) A circle needs a sideways force. There is none.

(4) A helix needs part of the velocity across the field. There is none.

(5) No force, so the velocity stays $v$ along the axis. This matches.

Why the other options fail

The force is zero, so there is no force along the axis to speed the proton up.

The force is zero, so there is no force along the axis to slow the proton down.

A circle needs a sideways force. The force is zero when the charge moves along the field.

A helix needs part of the velocity across the field. The proton has none, because it moves along the field.

Checks

$b * q * v * s$

Conclusion

$F = BqV\sin 0^\circ = 0$

No force acts on the proton, so it continues to move with velocity $v$ along the axis.

The answer is (5).

Ask first: what is the angle between the velocity and the magnetic field?

Put it in $F = BqV\sin\theta$. Along the field the force is zero, so the charge keeps its velocity.

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