Question

A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic field of magnitude 0.10 T . When the field is oriented normal to the slab's rectangular face, a Hall emf of 20 mV is measured across the slab's width. The probe is then placed in a magnetic field of unknown magnitude B, and a Hall emf of 69 mV is measured. Determine B assuming that the angle θ between the unknown field and the plane of the slab's rectangular face is

(a) θ = 90
(b) θ = 60

Answer 1

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf,
`V_(Hall) = 20\ mV = 0.02\ V`

Magnetic Field, B = 0.10 T

Hall emf,
`V'_(Hall) = 69\ mV = 0.069\ V`

Now,

Drift velocity,
`v_(d) = (V_(Hall))/(B)`

`v_(d) = (0.02)/(0.10) = 0.2\ m/s`

Now, the expression for the electric field is given by:

`E_(Hall) = Bv_(d)sin\theta` (1)

And

`E_(Hall) = V_(Hall)d`

Thus eqn (1) becomes

`V_(Hall)d = dBv_(d)sin\theta`

where

d = distance

`B = (V_(Hall))/(v_(d)sin\theta)` (2)

(a) When
`\theta = 90^(\circ)`

`B = (0.069)/(0.2* sin90) = 0.345\ T`

(b) When
`\theta = 60^(\circ)`

`B = (0.069)/(0.2* sin60) = 0.398\ T`