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Ivahew [28]
3 years ago
14

If i = 1.70 A of current flows through the loop and the loop experiences a torque of magnitude 0.0760 N ⋅ m , what are the lengt

hs of the sides s of the square loop, in centimeters?
Physics
2 answers:
vova2212 [387]3 years ago
8 0

The question is incomplete! The complete question along with answer and explanation is provided below

Question:

A wire loop with 40 turns is formed into a square with sides of length s. The loop is in the presence of a 2.0 T uniform magnetic field B that points in the negative y direction. The plane of the loop is tilted off the x-axis by 15°. If 1.70 A of current flows through the loop and the loop experiences a torque of magnitude 0.0760 N.m, what are the lengths of the sides s of the square loop, in centimeters?

Given Information:

Number of turns = N = 40 turns

Torque = τ = 0.0760 N.m

Current = I = 1.70 A

Magnetic field = B =  2 T

θ = 15°

Required Information:

Length of the sides of square loop = s = ?

Answer:

s = 4.64 cm

Explanation:

τ = NIABsin(θ)

Where N is the number of turns, I is the current flowing through the square loop, A is the area of square loop, B is the magnetic field, and θ is the angle between square loop and magnetic field strength with respect to the x-axis.

Re-arranging the equation to find out A

A = τ/ NIBsin(θ)

A = 0.0760/40*1.70*2*sin(15)

A = 0.00215 m²

We know that area of a square is

A = s²

Taking square root on both sides yields

s = √A

s = √0.00215 = 0.0464 m

in centimeters

s = 4.64 cm

Alik [6]3 years ago
6 0

Answer:

Length of the sides of the square loop is given by

s = √[(τ)/(NIB sin θ)]

Explanation:

The torque, τ, experienced by a square loop of area, A, with N number of turns around the loop and current of I flowing in the wire, with a magnetic field presence, B, and the plane of the loop tilted at angle θ to the x-axis, is given by

τ = (N)(I)(A)(B) sin θ

If everything else is given, the length of a side of the square loop, s, can be obtained from its Area, A.

A = s²

τ = (N)(I)(A)(B) sin θ

A = (τ)/(NIB sin θ)

s² = (τ)/(NIB sin θ)

s = √[(τ)/(NIB sin θ)]

In this question, τ = 0.076 N.m, I = 1.70 A

But we still need the following to obtain a numerical value for the length of a side of the square loop.

N = number of turnsof wire around the loop

B = magnetic field strength

θ = angle to which the plane of the loop is tilted, measured with respect to the x-axis.

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we know that, F = ma ------- let this be equation 2

so from equation 1 and 2

F = m( (v² - u²)/2s )

where m is mass, a is acceleration, u is initial velocity, v is final velocity and s is the displacement.

1)

Wearing sit belt, car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance.

i.e, m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 m

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F = 70( ((0)² - (18)²) / 2 × 1.1 )

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Therefore, the net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

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