A 2.0 meter length of straight conducting wire carries a current of 10.0A A.) Fi

Question

A 2.0 meter length of straight conducting wire carries a current of 10.0A

A.) Find the resulting magnetic field at a distance of 10.0 cm from the wire.

B.) If the wire is now wrapped to form a circular coil with a 10.0 cm radius, find the value of the resulting magnetic field at the center of the coil. (Note that you will need to compute the number of loops N that can be made from a 2.0m length of wire.)

C.) Compare the strength of the magnetic field for parts a and b and compare their ratio Bb/Ba In other words, how many times stronger is the field for the coil?

Explanation / Answer

a)The magnetic field due to a current carrying wire is given by:

B = u0 I/ 2 pi R

B = 4 pi x 10^-7 x 10 / 2 pi x 0.1 = 2 x 10^-5 T

Hence, B = 2 x 10^-5 T

b)Number if loops that can be formed will be:

C = n (2 pi r)

n = C/(2 pi r) = 2/ 2 x 3.14 x 0.1 = 3.2

The magnetic field at the center will be:

B = u0 n I/ 2 r

B = 4 pi x 10^-7 x 3.2 x 10 / 2 x 0.1 = 2.01 x 10^-4 T

Hence, B = 2.01 x 10^-4 T

Compare the strength of the magnetic field for parts a and b and compare their ratio Bb/Ba In other words, how many times stronger is the field for the coil?

c)Bb/Ba = 2.01 x 10^-4/2 x 10^-5 = 10

Hence, Bb/Ba = 10

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