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.

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 90^o. 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 3^rd law discussed {From Newton’s 3^rd 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