CAIE A-Level Physics 20 Magnetic Fields
Practise analysing magnetic fields, forces on conductors and charges, Hall and velocity-selection applications, magnetic flux and electromagnetic induction.
- Syllabus
- 2028–2030
- Course
- Physics 9702
- Level
- A2
Practise analysing magnetic fields, forces on conductors and charges, Hall and velocity-selection applications, magnetic flux and electromagnetic induction.
State what is meant by a magnetic field.
region where a force acts on
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a current-carrying conductor
or
a moving charge
or
a magnetic material / magnetic pole
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A long, straight wire P carries a current into the page, as shown in Fig. 6.1.

Fig. 6.1
On Fig. 6.1, draw four field lines to represent the magnetic field around wire P due to the current in the wire.
concentric circles around the wire
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spacing between circles increases with distance from wire
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arrows showing direction of field is clockwise
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A second long, straight wire Q, carrying a current of 5.0 A out of the page, is placed parallel to wire P , as shown in Fig. 6.2.

Fig. 6.2
The flux density of the magnetic field at wire Q due to the current in wire P is 2.6 mT .
Calculate the magnetic force per unit length exerted on wire Q by wire P.
force per unit length = Nm−1
F=B I L
C1
force per unit length = BI
=2.6×10−3×5.0=0.013 N m−1
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State the direction of the force exerted on wire Q by wire P.
to the right
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The flux density of the magnetic field at wire P due to the current in wire Q is 1.5 mT . Determine the magnitude of the current in wire P. Explain your reasoning.
current = A
force (per unit length) has the same magnitude due to Newton's 3rd law
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0.013=1.5×10−3×I
current =8.7 A
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Define magnetic flux density.
force per unit current
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force per unit length
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current / wire is perpendicular to (magnetic) field (lines)
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An insulated rectangular coil of wire, consisting of 40 turns, is suspended in a cradle from a newton meter, as shown in Fig. 7.1.

Fig. 7.1
The vertical sides of the coil have a length of 5.00 cm and the horizontal sides have a length of 3.00 cm . The initial reading on the newton meter is 0.563 N .
A U-shaped magnet rests on a top-pan balance that is set to a reading of 0.00 g . The lower edge of the coil is lowered into the region between the poles of the U-shaped magnet, as shown in the side view in Fig. 7.2.

Fig. 7.2
The magnetic field in the region between the poles is uniform.
The lower edge of the coil is entirely within the uniform magnetic field.
A current of 3.94 A is now passed through the coil. This causes the reading on the top-pan balance to change to 2.16 g .
Explain why the current causes a vertical force to act on the coil.
current (in coil) is perpendicular to magnetic field (so force on wire)
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force (on wire) is perpendicular to current and field (so is vertical)
or
current and field are both horizontal (so force is vertical)
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Determine, to three significant figures, the flux density B of the uniform magnetic field.
NBIL = mg
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B=(2.16×10−3×9.81)/(40×3.94×0.0300)
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=4.48×10−3 T
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Determine what is now the reading on the newton meter. Explain your reasoning.
(magnetic) forces (on balance and newton meter) are (equal and) opposite
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reading =0.563−(2.16×10−3×9.81)=0.542 N
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A uniform magnetic field is now applied in the region of the electric field in Fig. 5.1, so that the electron in (b)(iii) travels undeviated through the region.
Determine the direction of the uniform magnetic field.
into the page
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Explain, with reference to the forces exerted by the two fields on the electron, why the path of the electron is undeviated.
forces are in opposite directions
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(undeviated) when (magnitudes of) forces are equal
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Determine the flux density B of the uniform magnetic field. Give a unit with your answer.
B= unit
E q=B q v
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B=E/v=(6.4×103)/(2.6×107)=2.5×10−4 T
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