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Ira Lisetskai [31]
3 years ago
14

A small circular coil of 5 turns of wire lies in a uniform magnetic field of 0.8 T, so that the normal to the plane of the coil

makes an angle of 100◦ with the direction of B~ . The radius of the coil is 4 cm, and it carries a current of 1 A
Physics
1 answer:
Travka [436]3 years ago
8 0

Complete question:

A small circular coil of 5 turns of wire lies in a uniform magnetic field of 0.8 T, so that the normal to the plane of the coil makes an angle of 100◦ with the direction of B~ . The radius of the coil is 4 cm, and it carries a current of 1 A.

What is magnitude of the magnetic moment of the coil? Answer in units of A · m2.

Answer:

The magnetic moment of the coil is 0.0252 A.m²

Explanation:

Given;

radius of the coil, r = 4 cm = 0.04 m

number of turns of the coil, N = 5 turns

magnetic field strength B = 0.8 T

current in the coil, I = 1 A

Area of the coil, A = πr² = π(0.04)² = 0.00503 m²

magnetic moment of the coil, μ = NIA

where;

N is the number of turns

I is the current in the coil

A is the area of the coil

magnetic moment of the coil, μ = 5 x 1 x 0.00503 = 0.0252 A.m²

Therefore, the magnetic moment of the coil is 0.0252 A.m²

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Answer:

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Explanation:

The fundamental frequency of a standing wave on a string is given by

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}

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For the string in the problem,

L = 30.0 m

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Substituting into the equation, we find the fundamental frequency:

f=\frac{1}{2(30.0)}\sqrt{\frac{20.0}{(9.00\cdot 10^{-3}}}=0.786 Hz

The next frequencies (harmonics) are given by

f_n = nf

with n being an integer number and f being the fundamental frequency.

So we get:

f_2 = 2 (0.786 Hz)=1.572 Hz

f_3 = 3 (0.786 Hz)=2.358 Hz

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4 years ago
The distance between two stations is 180 km. A train takes 2 hours to cover this distance. The speed of the train in m/sec is...
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Av = 25 [m/s]

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