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## A length of conducting wire is bent into a circular coil of 7 turns. The diameter of the circle is 30.0 cm, and the axis of the coil is perp

Question

A length of conducting wire is bent into a circular coil of 7 turns. The diameter of the circle is 30.0 cm, and the axis of the coil is perpendicular to the horizontal component of the Earth’s magnetic field. A horizontally-oriented compass is placed along the axis of the coil, directly in the center, and when a current of 0.540 A flows in the coil, the compass needle is deflected by 45Â° from magnetic north.

Required:

a. What is the horizontal component of theEarth’s magnetic field?

b. The current in the coil is switched off. A “dip needle” is amagnetic compass mounted so that it can rotate in a verticalnorth-south plane. At this location, a dip needle makes an angle of11.0Â° from the vertical. What isthe total magnitude of the Earth’s magnetic field at thislocation?

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Physics
3 years
2021-09-04T16:31:45+00:00
2021-09-04T16:31:45+00:00 1 Answers
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## Answers ( )

Answer:a. 5.09 × 10⁻⁷ T b. 2.67 × 10⁻⁶ T

Explanation:a. What is the horizontal component of the Earth’s magnetic field?The magnetic field due to the circular coil, B = μ₀Ni/L where μ₀ = 4π × 10⁻⁷ H/m, N = number of turns of coil = 7, i = current = 0.540 A and L = length of coil = πND where N = number of turns of coil = 7 and D = diameter of coil = 30.0 cm = 0.30 m.

So, B = μ₀Ni/L

= μ₀Ni/πND

= μ₀i/πD

Substituting the values of the variables into the equation, we have

B = μ₀i/πD

= 4π × 10⁻⁷ H/m × 0.540 A/(π × 0.30 m)

= 4 × 10⁻⁷ H/m × 0.540 A/0.30 m

= 2.16 × 10⁻⁷ HA/m/0.30 m

= 7.2 × 10⁻⁷ T

Since this magnetic field is at 45° to the horizontal, its horizontal comoonent equals the horizontal component B’ of the earth’s magnetic field.

So, Bcos45° = B’

B’ = (7.2 × 10⁻⁷ T)cos45°

B’ = 5.09 × 10⁻⁷ T

b. What is the total magnitude of the Earth’s magnetic field at this location?Since the angle of dip at this location is 11.0° and B” is the magnetic field at this location, the horizontal component of B” = B’

So B”sin11.0° = 5.09 × 10⁻⁷ T

B” = 5.09 × 10⁻⁷ T/sin11.0°

B” = 5.09 × 10⁻⁷ T/0.1908

B” = 26.68 × 10⁻⁷ T

B” = 2.668 × 10⁻⁶ T

B” ≅ 2.67 × 10⁻⁶ T