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Hitman42 [59]
3 years ago
14

12. A flat circular coil of wire having 200 turns and diameter 6.0 cm carries a current of 7.0 A. It is placed in a magnetic fie

ld of with the plane of the coil making an angle of 30° with the magnetic field. What is the magnitude of the magnetic torque on the coil?
Physics
1 answer:
IgorLugansk [536]3 years ago
4 0

Answer:

The magnitude of the magnetic torque on the coil is 1.98 A.m²

Explanation:

Magnitude of magnetic torque in a flat circular coil is given as;

τ = NIASinθ

where;

N is the number of turns of the coil

I is the current in the coil

A is the area of the coil

θ is the angle of inclination of the coil and magnetic field

Given'

Number of turns, N = 200

Current, I = 7.0 A

Angle of inclination, θ = 30°

Diameter, d = 6 cm = 0.06 m

A = πd²/4 = π(0.06)²/4  = 0.002828 m²

τ = NIASinθ

τ = 200 x 7 x 0.002828 x Sin30

τ = 1.98 A.m²

Therefore, the magnitude of the magnetic torque on the coil is 1.98 A.m²

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Answer:

E=0 at r < R;

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E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}} at r >= 2R

Explanation:

Since we have a spherically symmetric system of charged bodies, the best approach is to use Guass' Theorem which is given by,

\int {E} \, dA=\frac{Q_{enclosed}}{\epsilon} (integral over a closed surface)

where,

E = Electric field

Q_{enclosed} = charged enclosed within the closed surface

\epsilon = permittivity of free space

Now, looking at the system we can say that a sphere(concentric with the conducting and non-conducting spheres) would be the best choice of a Gaussian surface. Let the radius of the sphere be r .

at r < R,

Q_{enclosed} = 0 and hence E = 0 (since the sphere is conducting, all the charges get repelled towards the surface)

at 2R > r > R,

Q_{enclosed} = Q,

therefore,

E\times4\pi r^{2}=\frac{Q_{enclosed}}{\epsilon}      

(Since the system is spherically symmetric, E is constant at any given r and so we have taken it out of the integral. Also, the surface integral of a sphere gives us the area of a sphere which is equal to 4\pi r^{2})

or, E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}

at r >= 2R

Q_{enclosed} = 2Q

Hence, by similar calculations, we get,

E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}

4 0
4 years ago
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Answer:

Option c is correct

Explanation:

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