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# 4 Magnetism problems: Magnitude, direction, tornado field, spectrometer, bloodflow

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1. At a certain location, Earth has a magnetic field of 0.6*10^4T pointing 75 degrees below the horixontal in a north-south plane. A 10m long straight wire carries a 15A current.
A) If the current is directed horizontally toward the east, what are the magnitude and direction of the magnetic force on the wire?
B) What are the magnitude and direction of the force if the current is directed vertically upward?

2. In 1962, measurements of the magnetic field of a large tornado were made at the Geophysical Observatory in Tulsa, Ok. If the tornado's field was B=1.5*10^-8T pointing north when the tornado was 9.00km east of the observatory, what current was carried up or down the funnel of the tornado? Model the vortex as a long straight wire carrying a current.

3. Consider the mass spectrometer shown schematically in the first attachment or Figure P19.30 cicled on the attachment, This is #30 not 36....At top of page.
The electric field between the plates of the velocity selector is 950V/m, and the magnetic fields in both the velocity selector and the deflection chamber have magnitudes of 0.930T. Calculate the radius of the path in the system for a singly-charged ion with mass m=2.18*10^-26kg.

4. A heart surgeon monitors the flow rate of blood through an artery using an electromagnetic flowmeter(shown in the second attachment #57 and it is circled)
Electrodes a and b make contact with the outer surface of the blood vessel, which has interior diameter3.00mm.
a). For a magnetic field magnitude of 0.0400T, a potential difference of 160uV apperars between the electrodes. Calculate the speed of the blood.
b). Verify that electrode a is positive, as shown. Does the sign of the emf depend on whether the mobile ions in the blood are predominantly positively or negatively charged? Explain?

##### Solution Summary

With excellent explanations, the problems are carefully solved

##### Solution Preview

See attached graph.

1. A) Magnetic force on a current carrying conductor is given by

F = L (I x B) Where I and B are vectors (I cross B ) L is the
length of the wire. Direction of the force will be perpendicular to the plane containing I and B.

F = 10 x 15 x 0.6 x10^4 Sin(75) See figure 1 also
= 86.93 x 10^4 Newton (Directed vertically upwards from the plane containing I and B)

B) The angle between I and B in this case is 90 deg; B in the plane of the paper and I perpendicular to it.

F = 10 x 15 x 0.6 x10^4 Sin(90) = 90 x 10^4 Newton
Now the direction of the force is pointing towards west, perpendicular to I and B. Use right hand thumb rule.
(Note: Check whether B is .6 x 10^4 or .6 x 10^-4 tesla)

2)
Consider it as a current carrying wire at a distance 9 km from a detector. We have to find the current that caused the detected field of 1.5 x 10^-8 T

The formula for the magnetic field due to a long current carrying straight conductor is,

B = [(mu0) I]/(2 pi d) where mu0 = 4pi x 10^-7 N/A^2 and d= 9km

Therefore, I = (2 B pi d)/(mu0)

= (2 x 1.5 x 10^-8 x 3.14 x 9 x 10^3)/(4 x 3.14 x 10^-7)
= 675 ...

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###### Education
• MSc , Pune University, India
• PhD (IP), Pune University, India
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