Question 1
(a) A straight conductor carrying a current I is
at an angle θ to a uniform magnetic field of flux density B, as shown in Fig.
6.1.
Fig. 6.1
The conductor and the magnetic field
are both in the plane of the paper. State
(i) an expression for the force per
unit length acting on the conductor due to the magnetic field, [1]
(ii) the direction of the force on
the conductor. [1]
(b) A coil of wire consisting of two loops is
suspended from a fixed point as shown in Fig. 6.2.
Fig. 6.2
Each loop of wire has diameter 9.4
cm and the separation of the loops is 0.75 cm.
The coil is connected into a circuit
such that the lower end of the coil is free to move.
(i) Explain why, when a current is
switched on in the coil, the separation of the loops of the coil decreases. [4]
(ii) Each loop of the coil may be
considered as being a long straight wire.
In SI units, the magnetic flux
density B at a distance x from a long straight wire carrying a current I is
given by the expression
B = (2.0 × 10–7) I / x
When the current in the coil is
switched on, a mass of 0.26 g is hung from the free end of the coil in order to
return the loops of the coil to their original separation. Calculate the
current in the coil. [4]
Reference: Past Exam Paper – November 2007 Paper 4 Q6
Solution 1:
(a)
(i) Force per unit length = BI sinθ
(ii) Direction: (downwards) into
(the plane of) the paper
{Fleming’s left hand rule:
Thumb = Force = ???, Forefinger = Field and Middle finger = Current = to right
here.
Note that in this case, B
and I are in same plane and the angle between them is NOT 90o. The force
SHOULD be perpendicular to both B and I.}
(b)
(i)
{Direction of the magnetic
field is obtained from the right hand grip rule}
EITHER
The magnetic field (due to the
current) in one loop cuts / is normal to the current in the second, causing a
force on the second loop. EITHER
Newton’s 3rd law discussed {From Newton’s 3rd
law, a force will also act on the first loop and gives rise to an attraction
between the 2 loops} OR vice versa clear gives rise to
attraction
OR
Each loop acts as a coil which
produces magnetic field. The fields are in the same direction {as the current in the 2 loops is in the same direction},
so the loops attract {causing the separation between
them to decrease}.
(ii)
F = BIL
Magnetic flux density, B = (2.0×10-7)
I / (0.75×10-2) {= 2.67×10-5 I}
Force F = (Weight = mg =) (0.26×10-3)
× 9.81 (= 2.55×10-3 N)
{length of one loop, L = circumference
= 2π r = 2π (d/2) = 2π (9.4×10-2) / 2 = 2π (4.7×10-2)}
2.55×10-3 = [(2.67×10-5)
× I] × I × [2π (4.7×10-2)]
Current I = 18 A
