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Alona [7]
3 years ago
10

A 0.6 m length of current carrying wire is placed in a uniform magnetic field oriented perpendicularly to the wire. The current

is 4 A and the magnitude of the magnetic field is 10 T. What is the force on the wire?
a. 0 N.
b. 2.4 N.
c. 6 N.
d. 24 N.
e. 30 N.
Physics
1 answer:
viktelen [127]3 years ago
3 0

Answer:

D. The force on the wire is 24 N.

Explanation:

Given;

length of the wire,  L = 0.6 m

current in the wire, I = 4 A

magnitude of the magnetic field, B = 10 T

The fore on the wire is calculated as;

F = BILsinθ

where;

θ is the angle of inclination of the wire on magnetic field = 90°

F = (10)(4)(0.6)(sin 90°)

F = 24 N

Therefore, the force on the wire is 24 N.

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Let the angle be Θ (theta)

Let the mass of the crate be m.

a) When the crate just begins to slip. At that moment the net force will be equal to zero and the static friction will be at the maximum vale.

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μ (coefficient of static friction) = 0.29

Static friction = μN = μmg CosΘ

Now, along the ramp, the equation of net force will be:

mg SinΘ - μmg CosΘ = 0

mg SinΘ = μmg CosΘ

tan Θ = μ

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Θ = 16.17°

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Coefficient of kinetic friction = μ = 0.26

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Plugging the values

a = 9.8 × 0.278 - 0.26 × 9.8 × 0.96

a = 2.7244 - 2.44608

a = 0.278 m/s^2

Hence, the acceleration is 0.278 m/s^2

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Part 1: Fill in the SI unit for each of the following measurements. 1. Time: 2. Length: 3. Mass: 4. Temperature:
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MrMuchimi

Answer:

Explanation:

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Bottom diagram: starts off large and decreases as we move from left to right  = - acceleration.  

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