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Vinil7 [7]
3 years ago
12

A current-carrying wire is bent into a circular loop of radius R and lies in an xy plane. A uniform external magnetic field B in

the +z direction exists throughout the plane of the loop. The current has the magnitude of I and it is deirected counterclockwise when observing from positive z axis.What is the magnetic force exerted by the external field on the loop?
Express your answer in terms of some or all of the variables I, R, and B
Physics
1 answer:
Ilya [14]3 years ago
8 0

Answer:

 F = 2π I R B

Explanation:

The magnetic force is described by the equation.

      F = q v x B = i L  x B

Where i is the current, L is a vector that points in the direction of the current (length) and B is the magnetic field.

This equation can be used in scalar form and the direction of the force found by the right hand ruler, the thumb goes in the direction of L, the fingers extended in the direction of B and the palm of the hand indicates the direction of the force if the load is positive

     F = i L B sin θ

In this case the wire is in the xy plane and the z-axis field whereby they are perpendicular, θ = 90º and sin 90 = 1

     F = i L B

The loop length is

    L = 2π R

    F = i 2π R B

    F = 2π I R B

The force is in the loop

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Answer:

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Explanation:

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Magnitude of the normal force  = ?

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A ray of light traveling through air strikes a piece of diamond at an angle of incidence equal to 56 degrees. Calculate the angu
Montano1993 [528]

Answer:

The angle of separation is  \Delta \theta =  0.93 ^o

Explanation:

From the question we are told that

    The angle of incidence is  \theta  _ i  = 56^o

     The refractive index of violet light  in diamond  is  n_v = 2.46

       The refractive index of red light in diamond is n_r = 2.41

      The wavelength of violet light is  \lambda _v = 400nm = 400*10^{-9}m

         The wavelength of red  light is  \lambda _r = 700nm = 700*10^{-9}m

Snell's  Law can be represented mathematically as

         \frac{sin \theta_i}{sin \theta_r} = n

Where \theta_r is the angle of refraction

=>       sin \theta_r  =   \frac{sin \theta_i}{n}

Now considering violet light

               sin \theta_r__{v}}  =   \frac{sin \theta_i}{n_v}

substituting values

                sin \theta_r__{v}}  =   \frac{sin (56)}{2.46}

                 sin \theta_r__{v}}  =  0.337

                 \theta_r__{v}}  =  sin ^{-1} (0.337)

                 \theta_r__{v}}  =  19.69^o

Now considering red light

               sin \theta_r__{R}}  =   \frac{sin \theta_i}{n_r}

substituting values

                sin \theta_r__{R}}  =   \frac{sin (56)}{2.41}

                 sin \theta_r__{R}}  =  0.344

                 \theta_r__{R}}  =  sin ^{-1} (0.344)

                 \theta_r__{R}}  = 20.12^o

The angle of separation between the red light and the violet light is mathematically evaluated as

                  \Delta \theta = \theta_r__{R}} -  \theta_r__{V}}

substituting values

                  \Delta \theta =20.12 - 19.69

                  \Delta \theta =  0.93 ^o

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