IBDP Physics 5.4 – Magnetic effects of electric currents: IB style Question Bank SL Paper 1

Question

A long straight vertical conductor carries a current I upwards. An electron moves with horizontal speed v to the right.

                           

What is the direction of the magnetic force on the electron?

A Downwards

B Upwards

C Into the page

D Out of the page

Answer/Explanation

Ans: A

\(\vec F_m=q(\vec v \times \vec B)\)
Now \(q = -e\)
\(\vec v = v \; i , \vec B = B \; j\)
\(\vec F_m=-evB(\vec i \times \vec j) =evB (-\vec k)\)

Question

An electron enters the region between two charged parallel plates initially moving parallel to the plates.

M18/4/PHYSI/SPM/ENG/TZ2/20

The electromagnetic force acting on the electron

A.     causes the electron to decrease its horizontal speed.

B.     causes the electron to increase its horizontal speed.

C.     is parallel to the field lines and in the opposite direction to them.

D.     is perpendicular to the field direction.

Answer/Explanation

Markscheme

C

\(\vec F_q=q\vec E\)
\(q= -e\)
\(\vec F_e =-e \vec E\)
opposite to Electric field

Question

A liquid that contains negative charge carriers is flowing through a square pipe with sides A, B, C and D. A magnetic field acts in the direction shown across the pipe.

On which side of the pipe does negative charge accumulate?

M18/4/PHYSI/SPM/ENG/TZ1/19

Answer/Explanation

Markscheme

A

\(\vec B=-B\;j\)
charge \(= -q\)
\(\vec v = v\; i\)
\(\vec F_m = q(\vec V \times \vec B)\)
\(=-q (v\; i \times -B\;j) =qvB(i \times j) =qvB \; k\)

A is in \(k\) direction

Question

The diagram shows two current-carrying wires, P and Q, that both lie in the plane of the paper. The arrows show the conventional current direction in the wires.

The electromagnetic force on Q is in the same plane as that of the wires. What is the direction of the electromagnetic force acting on Q?

Answer/Explanation

Markscheme

A

Due to current-carrying wires, P magnetic field developed at Q is in direction of inside the plan i.e. \(j\)

Current in current-carrying wires, Q is in \(i\) direction 

Now force on current-carrying wires, Q due to P is given as

\(\vec F_m = I (\vec l \times \vec B) = IlB (i \times j) =IlB \; k\)