Look at the photograph of the coil in the set-up section before procedure 1: cur

Question

Look at the photograph of the coil in the set-up section before procedure 1: current-carrying wires are wrapped around the square solenoid of side a, length L, and nturns per meter. Assume the current runs clockwise when you are looking down on the coil.

a) As shown in the photograph: at the geometric center of the cavity inside the solenoid: the magnetic field

o is approximately zero

o points away from you, into the page

o  points toward you, out of the page

o points to the right

o points to the left

o points upward

o points downward

b) Suppose we put current I through the coil, and the magnetic field at the center of the cavity is B. If we now double three quantities: a (the length of each side), n (the number of turns per meter) and L (the length of the coil), the new magnetic field at the center of the cavity will be

o 3B

o 2B    

o 4B

o B/2

o B

o B/3

o B/4

NOTE: In MKS units: T is Tesla, A is amperes, V is volts.
c) Using an equation for magnetic field: the units of ?o can be expressed as

o T-m/A

o A-m/T    

o A/T-m

o T-m/s

o T-s/m

o T-A/V

o A/T-s

o T-s/V

o T-A/m

o T-s/A

NOTE: ?o = 1.26x10-6 in MKS units (determined in question 1).
WATCH OUT FOR UNITS!!!

A square coil has n = 12 turns per cm, side a = 6.63 cm, and length L = 36.1 cm.

a) If the current through this coil is I = 80.4 A, find the magnitude of the magnetic field in the center of the coil.
B =   T

b) Suppose now a second coil has the same dimensions and current, except it is long enough to be considered "infinite". In this case, find the magnitude of the magnetic field in the center of this coil.
B =   T

c) Calculate the percent difference between the magnitude of B in the first and second coils.
percent difference = %

Explanation / Answer

1.

a) points away from you , into the page

b) B =oni

so answer is 2B

c) T-m/A

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