Question

what current has to be maintained in a circular coil of wire of 50 turns and 2.54 cm in radius in order to just cancel the effect of earths megnetic field at a place where the horizontal components of earths field is 1.86 × 10 -5 T​

Answer 1

Answer:0.015A

Explanation:

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Answer 2

Answer:

Current (I) = 146.04 A is needed to cancel out Earth's Horizontal Component of Magnetic field.

Explanation:

Provided that:

Earth's Horizontal Component of Magnetic field at this place (Bh) = 1.86 × 10^-5 T

Turns of the circular coil ( Solinoid) (N) = 50

Radius of the coil (r) = 2.54 cm = 0.0254 m

Let ,current through Solinoid (I) = I ampere

Magnetic Permittivity = μ0 = 4 × 10^-7 H/m

Cross sectional area of the coil (A) = π r^2

Area:

$A = π r^2 \\ = \pi \times {(0.0254)}^{2} \: {meter}^{2} \\ = 2.027 \times {10}^{ - 3} \: {meter}^{2}$

We know Magnatic field of Solinoid

(B) = μ.N.I .... equation (i)

Total Magnetic flux of Earth's Horizontal Component through the Solinoid;

ϕ = Bh . A ....equation (ii)

According to the given situation;

ϕ = Bh = 1.86 × 10^-5 T ... equation (iii)

Using these two equation we get:

$ϕ = B.A \\ B= μ \: N \: I \\ \\ so \: ϕ = (μ \: N \: I) \: . \: A \\ or \:I = \frac{ϕ}{(μ \: N \:A )} \\ I = \frac{1.86 \times {10}^{ - 5} }{(4 \pi \times {10}^{ - 7} \times 50 \times 2.027 \times {10}^{ - 3} ) } \: ampere \\ = 146.04 \: ampere$

So current (I) = 146.04 A