Question

A long wire carrying I=1.3A of constant current oriented East-to-West (and the current is running eastwards) is placed distance d=0.8cm over the top of a compass needle. Assuming the Earth magnetic field to be BEarth= 29μT what is the absolute value of the angular deviation θ of the compass needle from the North-to-South direction?

Answer 1

Answer:

Explanation:

Magnetic field due to a long wire

= 10⁻⁷ x 2i / r where i is current in the wire and r is distance of the point from the wire

In the given case magnetic field

B = 10⁻⁷ x 2i / r

= 10⁻⁷ x 2x 1.3 / .8 x 10⁻²

= 3.25 x 10⁻⁵ T

= 32.5 x 10⁻⁶ T

It will create a deflecting torque on the compass needle

Let after turning by angle θ , equilibrium point is reached

at this point , deflecting torque = restoring torque by earth's magnetic field

MB cosθ = MHsinθ , M is magnetic moment of magnetic needle , H is horizontal component of earth's magnetic field.

Tanθ = B / H

= 32.5 / 29

=1.20

θ = 50 degree approx .

Answer 2

The absolute value of the angular deviation θ of the compass needle from the North-to-South direction is; θ = 58.39°

We are given;

Current = 1.3 A

Distance; r = 0.8 cm = 0.008 m

Magnetic field = 20μT = 20 × 10⁻⁶ T

Formula for Magnetic field is;

B = μ_o*I/(2πr)

Where;

μ_o = 4π × 10⁻⁷ H/m

Plugging in the relevant values;

B = (4π × 10⁻⁷ × 1.3)/(2π × 0.008)

B = 32.5 μT

Formula for the angular deviation θ is;

θ = tan^(-1) 32.5/20

θ = 58.39°

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