Question

A 500-turn circular loop 15.0 cm in diameter is initially aligned so that its axis is parallel to the Earth's magnetic field. In 2.77 ms the coil is flipped so that its axis is perpendicular to the Earth's field. If a voltage of 0.166 V is induced in the coil, what is the value of the Earth's magnetic field?

Answer 1

the value of the earth's magnetic field is 2.60×10^-3 T.

Step-by-step explanation:

we know that:

ε = N×A×2×(ΔB/Δt)

ΔB = ε×Δt/2×N×A

Bf - Bi = ε×Δt/2×N×A

Bf -0 = (0.166)×(2.77×10^-3)/2×(500)×(π×(7.5×10^-3)^2)

Bf = 2.60×10^-3 T

Therefore, the value of the earth's magnetic field is 5.82×10^--5 T.

Answer 2

The magnetic field would be B = 5.2 x
`10^(-5)` T

Step-by-step explanation:

The voltage that is obtained when the magnetic flux of a coil is altered is the induced emf.

The induced emf E = rate of change of magnetic flux

When the loop axis is parallel to the Earth's magnetic field the initial flux

is;

φ1 = NBAcosθ

Final flux = 0

Therefore the change in flux is dφ = NBAcosθ - 0

dφ = NBAcosθ

Given

the induced emf E = 0.166 V

the time dt = 2.77 ms = 2.77 x
`10^(-3)` s

N = number of turns = 500

d = diameter = 15 cm /100 = 0.15 m

A = area of cross section = π
`d^(2)`/4 = π x (0.15m
`)^(2)`/4 = 0.0176715
`m^(2)`

B = magnetic field

The induced emf E can be expressed thus

E = dφ/dt

= NBAcosθ/dt

cosθ = 1

E =NBA/dt

0.166 = 500 x Bx 0.0176715/ 2.77 x
`10^(-3)`

0.166 x 2.77 x
`10^(-3)` = 500 x B x 0.0176715

0.00045982 = 8.83572934 B

B = 0.00045982 / 8.83572934

B = 0.00005204097

B = 5.2 x
`10^(-5)` T

Therefore the magnetic field would be B = 5.2 x
`10^(-5)` T