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Tatiana [17]
3 years ago
6

A loop of wire is placed in a uniform magnetic field. (a) For what orientation of the loop is the magnetic flux a maximum? The p

lane of the loop is perpendicular to the magnetic field. The plane of the loop is parallel to the magnetic field. The magnetic flux is independent of the orientation of the loop. (b) For what orientation is the flux zero? The plane of the loop is perpendicular to the magnetic field. The plane of the loop is parallel to the magnetic field. The magnetic flux is independent of the orientation of the loop.
Physics
1 answer:
ra1l [238]3 years ago
5 0

Answer:

1) The plane of the loop is perpendicular to the magnetic field.

2) The magnetic flux is independent of the orientation of the loop.p

Explanation:

The flux is calculated as φ=BAcosθ. The flux is therefore the highest when the magnetic field vector is perpendicular to the plane of the loop We can also deduce that the flux is zero when there is no magnetic field part perpendicular to the loop When the angle reaches zero, the flux is in the limit because when the angle becomes zero, the cos is the maximum.

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Answer: Option (1) is the correct answer.

Explanation:

Formula to calculate correlation coefficient is as follows.

           Correlation coefficient = \sqrt{R^{2}}

As the slope for given equation is positive so, it means that the correlation will also be positive in nature.

Therefore, we will calculate the value of correlation coefficient as follows.

         Correlation coefficient = \sqrt{R^{2}}

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                                               = 0.54

Thus, we can conclude that the correlation coefficient, r is 0.54.          

8 0
3 years ago
A charge −1.3 × 10−5 C is fixed on the x-axis at 7 m, and a charge 1 × 10−5 C is fixed on the y-axis at 4 m. Calculate the magni
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Answer:

6104 N/C.

Explanation:

Given:

k = 8.99 × 10^9 Nm2/C^2

Qx = 1.3 × 10^-5 C

rx = 7 m

Qy = 1 × 10−5 C

ry = 4 m

E = F/Q

= kQ/r^2

Ex = (8.99 × 10^9 × 1.3 × 10^−5) ÷ 7^2

= 2385.1 N/C.

Ey = (8.99 × 10^9 × 1.0 × 10^−5) ÷ 4^2

= 5618.75 N/C

Eo = sqrt(Ex^2 + Ey^2)

= sqrt(3.157 × 10^7 + 5.69 × 10^6)

= 6104 N/C.

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4 years ago
A 0.030 kg lead bullet hits a steel plate, both initially at 20?C. The bullet melts and splatters on impact. Assume that 80% of
Dmitry_Shevchenko [17]

Answer:

(a). The required is 1871.2 J.

(b). The speed the lead bullet is 395 m/s.

(c). The 20% of energy must have gone into collision between steel plate and bullet.

Explanation:

Given that,

Mass of bullet = 0.030 kg

Temperature = 20°C

(a). We need to heat required to increase the temperature of the lead bullet and melt it

Using formula of heat

Q=mS\Delta T+mL

Where, m = mass of lead bullet

S = specific heat

L = latent heat

T = Temperature

Put the value into the formula

Q=0.030\times130\times(327.5-20)+0.030\times22400

Q=1871.2\ J

The required is 1871.2 J.

(b). Assume that 80% of the bullet’s kinetic energy goes into increasing its temperature and then melting it

We need to calculate the energy

Q=\dfrac{1871.2}{0.8}

Q=2339 J

We need to calculate the speed the lead bullet

Using formula of speed

K.E=Q

\dfrac{1}{2}mv^2=2339

v^2=\dfrac{2339\times2}{0.030}

v=\sqrt{\dfrac{2339\times2}{0.030}}

v=394.8 = 395\ m/s

The speed the lead bullet is 395 m/s.

(c). The 20% of energy must have gone into collision between steel plate and bullet.

Hence, This is the required solution.

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3 years ago
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