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den301095 [7]
3 years ago
10

Compare the magnitude of the magnetic field at the center of a circular current loop of radius 30 mm with the magnitude of the m

agnetic field at the center of a solenoid of the same radius and with 3.0 turn per millimeter. Assume the current is the same through the current loop and the solenoid.
Physics
1 answer:
NISA [10]3 years ago
4 0

Answer:

Ratio of magnetic field will be \frac{B}{B'}=0.0055

Explanation:

We have given radius of the loop r = 30 mm = 0.03 m

We know that magnetic field at the center of the loop is given by

B=\frac{\mu _0i}{2r}---------eqn 1

Number of turns in the solenoid is given as n = 3 turn per mm = 3000 turn per meter

We know that magnetic field due to solenoid is given by

B'=n\mu _0i-------------eqn 2

Now dividing eqn 1 by eqn 2

\frac{B}{B'}=\frac{\frac{\mu _0i}{2r}}{\mu _0ni}=\frac{1}{2nr}=\frac{1}{2\times 3000\times 0.03}=0.0055

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Answer: The correct answer is- The plant with the smooth seeds is homozygous for the dominant trait.

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Let us consider 'S' depicts the allele for smooth seed and 's' for the wrinkled seeds. There can be two genotypes for smooth seeds that is 'Ss' and 'SS' whereas only one genotype is possible for wrinkled ( 'ss') seeds.

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