Q#14
A thin rectangular slice of aluminium has sides of length 65 mm, 50 mm and 0.10 mm, as shown in Fig. 9.1.
Fig. 9.1 (not to scale)
Some of the corners of the slice are labelled.
A current I of 3.8 A is normal to face RSXY of the slice.
In aluminium, the number of free electrons per unit volume is 6.0 × 1028 m−3.
A uniform magnetic field of magnetic flux density B equal to 0.13 T is normal to face QRYZ of the aluminium slice in the direction from Q to P.
A Hall voltage VH is developed across the slice and is given by the expression
VH = BI / ntq
.
(a) Use Fig. 9.1 to state the magnitude of the distance t. [1]
(b) Calculate the magnitude of the Hall voltage VH. [2]
[Total: 3]
Solution:
(a)
0.10 mm
{In the equation, t is the thickness of materials through which the current passes.}
(b)
{VH = BI / ntq
Thickness t should be in metres.
Charge of electrons: q = 1.60×10-19}
VH = (0.13 × 3.8) / (6.0×1028 × 0.10×10-3 × 1.60×10-19)
VH = 5.1 × 10-7 V
Post a Comment for "A thin rectangular slice of aluminium has sides of length 65 mm, 50 mm and 0.10 mm, as shown in "