Physics 9702 · AS & A Level · Force on a moving charge

Force on a moving charge — practice question

A particle with mass $m$ and charge $-q$ moves through a vacuum at constant speed $v$. It then enters a uniform magnetic field with flux density $B$. The angle at entry between the particle’s direction of travel and the magnetic-field direction is $90^\circ$.
(a)[3]

Explain why the particle’s path in the magnetic field is an arc of a circle.

(b)[1]

The radius of the arc in (a) is $r$. Show that the particle’s ratio $\frac{q}{m}$ is given by $\frac{q}{m} = \frac{v}{Br}$.

(c(i))[2]

The particle’s initial speed $v$ is $2.0 \times 10^7\,\text{m s}^{-1}$. The magnetic flux density $B$ is $2.5 \times 10^{-3}\,\text{T}$. The radius $r$ of the arc in the magnetic field is $4.5\,\text{cm}$. Use these data to calculate the ratio $\frac{q}{m}$.

(c(ii))[2]

The route taken by the negatively-charged particle before it enters the magnetic field is shown in Fig. 6.1. The direction of the magnetic field is into the plane of the paper. On Fig. 6.1, sketch the particle’s path while it is in the magnetic field and after it leaves the field.

(ii)[2]

On Fig. 6.1, sketch the particle’s path in the magnetic field and after it emerges from the field.

Worked solution & mark scheme

This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: Magnetic-field force remains constant

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