Saturday, February 4, 2017

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:
(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.}

{Direction of the magnetic field is obtained from the right hand grip rule}
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

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}.           

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

Saturday, January 28, 2017

A potentiometer circuit that is used as a means of comparing potential differences

Question 1
A potentiometer circuit that is used as a means of comparing potential differences is shown in Fig. 5.1.

Fig. 5.1
A cell of e.m.f. E1 and internal resistance r1 is connected in series with a resistor of resistance R1 and a uniform metal wire of total resistance R2.
A second cell of e.m.f. E2 and internal resistance r2 is connected in series with a sensitive ammeter and is then connected across the wire at BJ. The connection at J is halfway along the wire. The current directions are shown on Fig. 5.1.

(a) Use Kirchhoff’s laws to obtain the relation
(i) between the currents I1, I2 and I3, [1]

(ii) between E1, R1, R2, r1, I1 and I2 in loop HBJFGH, [1]

(iii) between E1, E2, r1, r2, R1, R2, I1 and I3 in the loop HBCDJFGH. [2]

(b) The connection at J is moved along the wire. Explain why the reading on the ammeter changes. [2]

Reference: Past Exam Paper – November 2011 Paper 22 Q5

Solution 1:
(i) I2 = I1 + I3   (Kirchoff’s 1st law: Sum of current entering a junction = sum of current leaving the junction)

{Put e.m.f of the cell on one side. Since a current I1 is coming from the +ve terminal of the cell, the same current should come back to (negative terminal of) the cell. So, current I1 flows through r1 and R1 which are in series.
At junction B, the current I1 changes into I2 and I3 (according to the above equation) but since we are considering loop HBJFGH, we can neglect I3. From B to J, the current I2 flows. Since J is halfway on the wire, half of the total resistance of the wire (R2) should be considered [since resistance is directly proportional to the length of wire –> R = ρL / A]. So, the p.d. across BJ = I2(R2/2).
Now, consider from the junction J to part F, then G up to the negative terminal of cell E1. There is no more junction where the current changes. So, the current going back to the back is the same as that flowing through part JF of the wire (that is, I1). And again, since J is halfway on the wire, part JF would correspond to a resistance of R2/2. So, the p.d. across JF = I1(R2/2). Hence, the equation below is obtained.}
E1 = I2(R2/2) + I1(R2/2) + I1R1 + I1r1  

{Tip: Put emf of cells (E1 and E2) on one sides of equation and potential differences in components on other side. (Kirchoff’s 2nd law)
E2 is in opposite direction of E1 (check direction of currents I3 and I1).
I2 not present since we are not considering BJ. Current coming from J towards F is I1 (corresponding to pd = I1R2/2). Current across r2 in JDCB is I3(corresponding to pd = I3r2). In GH, pd = I1R1 + I1r1.}
E1 – E2 = -I3r2 + I1(R1 + r1 + R2/2)

(b) The potential difference across BJ of the wire changes / resistance of BJ changes. There is a difference in potential difference across the wire and the potential difference across cell E2.                  
{Since V = IR, current I = V / R. As there is a difference in V, the current also changes}

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