Motion in electromagnetic fields

00:00 8 marks

A proton (charge +*e*, mass *m*_p) enters a region of uniform magnetic field *B* with velocity *v*. The velocity vector is perpendicular to the magnetic field direction.

Part a

1 minutes 1 marks

Question

State the shape of the path followed by the proton in the magnetic field. [1]

Hint Hint

Hint

Consider the direction of the magnetic force relative to the velocity.

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Complete Answer

Circular path / arc of a circle.

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Part b

3 minutes 3 marks

Question

Show that the radius *r* of the path is given by *r* = *m*_p*v* / (*eB*). [3]

Hint Hint

Hint

Equate the magnetic force acting on the proton to the centripetal force required for circular motion.

Answer Answer
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The derivation involves equating the magnetic force \(F_B = evB\) (since velocity is perpendicular to the field) to the centripetal force \(F_c = m_p v^2 / r\), leading to \(r = m_p v / (eB)\).

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Part c

2 minutes 2 marks

Question

An alpha particle (charge +2*e*, mass ≈ 4*m*_p) enters the same magnetic field with the same velocity *v*. Determine the ratio of the radius of the alpha particle’s path to the radius of the proton’s path (*r*_α / *r*_p). [2]

Hint Hint

Hint

Use the formula derived in part (b) for both the proton and the alpha particle, substituting the correct charge and mass for the alpha particle.

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\(r_{\alpha} / r_p = 2\)

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Part d

2 minutes 2 marks

Question

The magnetic field is now replaced by a uniform electric field *E* acting in the same direction as the original magnetic field (perpendicular to *v*). Describe the path of the proton in this electric field. [2]

Hint Hint

Hint

Consider the force exerted by the electric field. Is it constant or changing? How does its direction relate to the initial velocity?

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Complete Answer

The path is parabolic. The electric force \(F_e = eE\) acts perpendicularly to the initial velocity \(v\), causing constant acceleration in that direction, similar to projectile motion.

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