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Ksenya-84 [330]
3 years ago
11

A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie

ld 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
Physics
1 answer:
Citrus2011 [14]3 years ago
5 0

Answer:

(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} = \frac{V_{Hall}}{B}

v_{d} = \frac{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 = \frac{V_{Hall}}{v_{d}sin\theta}                      (2)

(a) When \theta = 90^{\circ}

B = \frac{0.069}{0.2\times sin90} = 0.345\ T

(b) When \theta = 60^{\circ}

B = \frac{0.069}{0.2\times sin60} = 0.398\ T

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Explanation:

From the question we are told that

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Generally from the law of  energy conservation we have that

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So

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Here a is the amplitude of the subsequent oscillations

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