# IBDP Physics 5.4 – Magnetic effects of electric currents: IB Style Question Bank HL Paper 1

### Question

An electron enters the space inside a current-carrying solenoid. The velocity of the electron is parallel to the solenoid’s axis. The electron is

A slowed down.

B speeded up.

C undeflected.

D deflected outwards.

Ans: A

This question is bit tricky.

If we consider solenoid to be long or ideal then

$$\vec F_m = q(\vec V \times \vec B)$$
$$q = -e$$  and  $$\vec V$$  and  $$\Vec B$$  are in parallel hence
$$\vec V \times \vec B = VB Sin \theta = 0 as \theta =0$$
and hence electron will go undeflected and Answer is C

But if we consider real solenoid of finite length then

There is magnetic field outside solenoid and this will have net force on electron , but that will be very small as you can see $$B$$ at point $$P_2$$ is very weak (sparse line).

This force will oppose electron movement.

### Question

The diagram shows the magnetic field surrounding two current-carrying metal wires P and Q. The wires are parallel to each other and at right angles to the plane of the page.

What is the direction of the electron flow in P and the direction of the electron flow in Q?

### Markscheme

B

The direction of the magnetic field at a point P due to a long, straight wire can be found by a slight variation in the right-hand thumb rule If we stretch the thumb of the right hand along the long current and curl our fingers to pass through the point P, the direction of the fingers at P gives the direction of the magnetic field there. Dot indicate coming out of the plane

Hence at point P current is out of plan. Now flow of electron is opposite to current in wire , hence direction of electron at P is into page and at Q Out of page.

### Question

The diagram shows the path of a particle in a region of uniform magnetic field. The field is directed into the plane of the page.

This particle could be

A. an alpha particle.

B. a beta particle.

C. a photon.

D. a neutron.

### Markscheme

A

$$\vec B= B(-k)$$
$$\vec v= vi$$
$$\vec F_m = q(\vec v \times \vec B)$$
$$\vec F_m = F_m j$$  as particle moves towards y direction
$$F_m j= q(vi \times B(-k) )=qvB (k \times i)$$
Now  $$k \times i = j$$
hence q has to be positive otherwise direction will reverse
Hence particle has to be charged and positive which is alpha $$(He^{++})$$
beta is negative charged , photon and neutron are neutral

### Question

An ion follows a circular path in a uniform magnetic field. Which single change decreases the radius of the path?

A. Increasing the mass of the ion

B. Increasing the charge of the ion

C. Increasing the speed of the ion

D. Decreasing the magnetic flux density of the field

### Markscheme

B

FORCE ACTING ON A CHARGED PARTICLE MOVING IN A UNIFORM MAGNETIC FIELD

The force acting on a particle having a charge q and moving with velocity  in a uniform magnetic field $$\vec B$$ is given by
,
where θ is the angle between $$\vec v$$ and $$\vec B$$
If θ = 90°, F = qvB
In this case the force acting on the particle is maximum and this force acts as centripetal force which makes the charged particle move in a circular path.
where r is the radius of the circular path.
Hence r will decrease when q increases.

### Question

A metal rod M is falling vertically within a horizontal magnetic field. The metal rod and magnetic field are directed into the paper. What is the direction of the initial force acting on the metal rod that is predicted by Lenz’s law?

### Markscheme

B

Given \vec v\ =v(-j)

\vec B = B (-k)

According to lenz law , current will be induced in rod due to movement of electron in s direction such that force will oppose its vertical movement. Hence force will be induced upward which will oppose it’s moving downward.

### Examiners report

The correct response here is B. As the rod falls, an electric current in induced, forcing the electrons across the diameter of the wire. This in turn, produces an upward force on the rod as a whole in line with Lenz’s Law.