the right edge of the circuit in the figure extends into a 50 mt uniform magnetic field.

Answers

Answer 1

The magnitude of the circuit is 2.5*10^-2. Since the circuit's right edge in the illustration extends into a 50 mt uniform magnetic field, Rightward is the direction of the magnetic force.

According to the left hand rule, the thumb of the left hand points in the direction of force when the middle finger and index finger point in the direction of current and magnetic field, respectively. The magnetic field and the downward direction are indicated by the way the current in this circuit flows, which is from top to bottom. The thumb points to the right if the index finger points into the page and the middle finger points downward. Thus, the direction of the magnetic force is towards right.

F = (5.0)(10)(50mT)(10^-3/1)(sin90)

F = 2.5*10^-2N

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The complete question is -

The Right Edge Of The Circuit In The Figure Extends Into A 50 MT Uniform Magnetic Field. What Are The Magnitude of the circuit.

Related Questions

Light rays from the sun make a 300 angle to the vertical when seen from below the surface of a body of water. At what angle above the horizon is the sun? The indices of refraction of glass, water, and air are 1.50, 1.333, and 1.000293, respectively.

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The light rays from the sun is making an angle of 30° then the angle above the horizon of the sun is 48.32°.

What is an angle?

When two lines converge at a point, they produce an angle. The word "angle" refers to the length of the "gap" between these two rays. The symbol is used to denote it.

Radians, a unit of roundness or rotation, and degrees are the 2 most common units used to measure angles. In daily life, angles are present.

As per the information given in the question,

Use the equation of snell's law,

(sin i)/(sin r) = n₂/n₁

r = sin⁻¹ (n₁ sin i)/n₂

Put n₁ = 1.33 and n₂ = 1 also substitute i = 30°

r = sin⁻¹ (1.33) (sin 30)/1

r = sin⁻¹(0.665)

r = 41.68°

The formula for the angle above the horizon,

θ = 90° - r

θ = 90° - 41.68°

θ = 48.32°.

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exitance is defined as the power per unit area that is emitted from the surface of an object. for a blackbody radiator, the exitance emitted is dependent upon its temperature. calculate the exitance in w/cm2 for a blackbody radiator at 46 k and 335 k. exitance at 46 k

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The exitance in W/cm² for a blackbody radiator at 335 K is 232 W/cm².

The exitance of a blackbody radiator is given by the Stefan-Boltzmann law: exitance = sigma × [tex]T^4[/tex]

where exitance is the power per unit area emitted from the surface of the object, sigma is the Stefan-Boltzmann constant (5.67 x [tex]10^{-8}[/tex] W/m² K^4), and T is the temperature of the blackbody radiator in Kelvin.

To calculate the exitance in W/cm² for a blackbody radiator at 46 K, we can substitute this temperature into the equation:

exitance = 5.67 x [tex]10^{-8}[/tex] W/m² [tex]K^4[/tex] × [tex](46 K)^4[/tex] = 0.000715 W/m²

Converting this to W/cm², we get:

exitance (W/cm²) = 0.000715 W/m² × (100 cm/m)² = 0.0715 W/cm²

To calculate the exitance in W/cm² for a blackbody radiator at 335 K, we can substitute this temperature into the equation:

exitance = 5.67 x [tex]10^{-8}[/tex] W/m² [tex]K^4[/tex] × [tex](335 K)^4[/tex] = 2.32 W/m²

Converting this to W/cm², we get:

exitance (W/cm²) = 2.32 W/m² × (100 cm/m)² = 232 W/cm²

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