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Airida [17]
3 years ago
6

A current-carrying wire 0.50 m long is positioned perpendicular to a uniform magnetic field. If the current is 10.0 A and there

is a resultant force of 3.0 N on the wire due to the interaction of the current and field, what is the magnetic field strength?
Physics
1 answer:
____ [38]3 years ago
7 0

<em>Answer</em>


0.6 teslas


<em>Explanation</em>

When a conductor is inside a magnetic field it experiences a force given by;

Force = ILBsinθ


Where I⇒ current

L ⇒length of the conductor

B ⇒ magnetic field strength

θ ⇒ Angle between the conductor and magnetic field.


F = ILBsinθ


When θ = 90°, Then sin 90 =1 and the formula becomes;

F =ILB

3 = 10 × 0.5 × B

3 = 5B

B = 3/5

= 0.6


magnetic field strength = 0.6 teslas


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A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind
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Part a)

F_v = 4.28 N

Part B)

L = 1.02 m

Part C)

v = 1.25 m/s

Explanation:

Part A)

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so we will have

Tcos\theta = mg

T sin\theta = F_v

\frac{F_v}{mg} = tan\theta

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F_v = 4.28 N

Part B)

Here we can use energy theorem to find the distance that it will move

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7 0
3 years ago
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