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natima [27]
3 years ago
5

A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1 ° from the normal to the plane of the l

oop. The magnitude of the magnetic field is 4.35 T , and the radius of the wire loop is 0.280 m . Find the magnetic flux Φ through the loop.
Physics
1 answer:
umka2103 [35]3 years ago
8 0

Answer:

Magnetic flux through the loop is 1.03 T m²

Explanation:

Given:

Magnetic field, B = 4.35 T

Radius of the circular loop, r = 0.280 m

Angle between circular loop and magnetic field, θ = 15.1⁰

Magnetic flux is determine by the relation:

\Phi = BA\cos \theta   ....(1)

Here A represents area of the circular loop.

Area of circular loop, A = πr²

Hence, the equation (1) becomes:

\Phi=B\pi r^{2} \cos \theta

Substitute the suitable values in the above equation.

\Phi=4.35\times\pi (0.28)^{2} \cos 15.1

\Phi = 1.03 T m²

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6 0
3 years ago
A man pulled a 13.0 kg object 11.8 cm vertically with his teeth. (a) How much work (in J) was done on the object by the man in t
jonny [76]

Answer:

(a)The work done by the man is -15.03J.

(b)The force exerted on the object is 127.4N.

Explanation:

Mass of the object pulled by the man is -13kg

Object is lifted 11.8 cm vertical with his teeth it means (displacement = +11.8cm = +0.118m)

Acceleration due to gravity is 9.8 \mathrm{m} / \mathrm{s}^{2}

(a) <u>Calculating the work done</u>:

Work done = mgh

Where "m" is mass of an object, "g" is acceleration due to gravity and "h" is the displacement.

\text { Work }=-13 \times 9.8 \times(+0.118 \mathrm{m})

\text { Work }=-15.03 \mathrm{J}

The work done by the man is -15.03J.

(b) <u>Calculating the force</u>:

Probably the man and the object are close to the exterior of the earth. If the rigidity required to maintained the object of consistent velocity interior the gravitational field of the earth is \mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}

Thus the weight of the object is balanced by the force of the man's teeth on the object. That is

F = mg

\mathrm{F}=13 \times 9.8

F = 127.4N

The force exerted on the object is 127.4N.

4 0
3 years ago
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