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3 years ago

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1. You are designing a new solenoid and experimenting with material for each turn. The particular turn you are working with is a

circular loop of radius 3.50 cm that carries a current of 15.0 A . Calculate the magnetic field B at the center of the loop.
2. Consider the coordinate system in the figure. If the circular loop lies in the xy plane and if the current flows clockwise around the loop, which direction does the magnetic field point inside the loop?
A. Positive x.
B. Positive z.
C. Negative y.
D. Negative z.
3. If we decrease the radius the magnetic field will:_______.
A. Increase.
B. Decrease.
C. Stay the same.
D. None of the above.

1 answer:

svetoff [14.1K]3 years ago

4 0

Answer:

Explanation:

1 ) Magnetic field due to a circular coil carrying current

= μ₀I / 2r

I is current , r is radius of the wire , μ₀ = 4π x 10⁻⁷

= 4π x 10⁻⁷ x 15 / (2 x 3.5 x 10⁻²)

= 26.9 x 10⁻⁵ T

2 )

Negative z direction .

The direction of magnetic field due to a circular coil having current is known

with the help of screw rule or right hand thumb rule.

3 )

If we decrease the radius the magnetic field will:__increase _____.

A. Increase.

Magnetic field due to a circular coil carrying current

B = μ₀I / 2 r

Here r is radius of the coil . If radius decreases magnetic field increases.

So magnetic field will increase.

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Two parallel wires carry currents in the same direction. If the currents in the wires are 1A and 4A and the wires are 5 m apart.

serious [3.7K]

Answer:

$1.6\times 10^{-7}$ N

$2.4\times 10^{-7}$ N

Explanation:

$i_{1}$ = 1 A

$i_{2}$ = 4 A

$r$ = distance between the two wire = 5 m

$F$ = Force per unit length acting between the two wires

Force per unit length acting between the two wires is given as

$F = \frac{\mu _{o}}{4\pi }\frac{2i_{1}i_{2}}{r}$

$F = (10^{-7})\frac{2(1)(4)}{5}$

$F = 1.6\times 10^{-7}$ N

$r'}$ = distance of each wire from the midpoint = 2.5 m

Magnetic field midway between the two wires is given as

$B = \frac{\mu _{o}}{4\pi } \left \left ( \frac{2i_{2}}{r'} \right - \frac{2i_{1}}{r'} \right \right ))$

$B = (10^{-7}) \left \left ( \frac{2(4)}{2.5} \right - \frac{2(1)}{2.5} \right \right ))$

$B = 2.4\times 10^{-7}$

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3 years ago

a moving ball rolls into a stationary ball. the total momentum of both balls after the collision will be

rosijanka [135]

The momentum of the rolling ball will have less momentum than before the collision and the stationary ball will have more momentum after the collusion.

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4 years ago

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A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and then the separation betw

TEA [102]

Answer:

The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)

Explanation:

The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. $Q$ , the amount of charge stored in this capacitor, will stay the same.

The formula $\displaystyle Q = C\, V$ relates the electric potential across a capacitor to:

$Q$ , the charge stored in the capacitor, and
$C$ , the capacitance of this capacitor.

While $Q$ stays the same, moving the two plates apart could affect the potential $V$ by changing the capacitance $C$ of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:

$\displaystyle C = \frac{\epsilon\, A}{d}$ ,

where

$\epsilon$ is the permittivity of the material between the two plates.
$A$ is the area of each of the two plates.
$d$ is the distance between the two plates.

Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of $\epsilon$ . Neither will that change the area of the two plates.

However, as $d$ (the distance between the two plates) increases, the value of $\displaystyle C = \frac{\epsilon\, A}{d}$ will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.

On the other hand, the formula $\displaystyle Q = C\, V$ can be rewritten as:

$V = \displaystyle \frac{Q}{C}$ .

The value of $Q$ (charge stored in this capacitor) stays the same. As the value of $C$ becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.

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3 years ago

A single conservative force Fx = (6.0x  12) N (x is in m) acts on a particle moving along the x axis. The potential energy asso

jasenka [17]

Answer:

$U(3)=-43J$

Explanation:

Potential energy is minus the integral of Fdx. Doing the integration yields:

$U=\int\limits {6.0x+12}\, dx$

$U=-3x^2-12x+C$

$U(0)=20J$

so

$U(0)=-3(0)^2+12(0)+C=20$

$C=20$

Now for x=3.0m

$U(3)=-3*(3)^2-12(3)+20$

$U(3)=-43J$

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3 years ago

A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and ro

Mkey [24]

Answer:

a. Rotational inertia: 5.21kgm²

b. Magnitude of it's angular momentum: 123.32kgm²/s

Explanation:

Length of the rod = 3.46m

Weight of the rod = 12.8 N

Angular velocity of the rod= 226 rev/min

a. Rotational Inertia (I) about its axis

The formula for rotational inertia =

I = (1/12×m×L²) + m × ( L ÷ 2)²

Where L = length of the rod

m = mass of the rod

Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.

Acceleration due to gravity = 9.81m/s²

Mass of the rod = 12.8N/ 9.81m/s²

Mass of the rod = 1.305kg

Rotational Inertia =

(1/12× 1.305 × 3.46²)+ 1.305 ( 3.46÷2)²

Rotational Inertia = 1.3019115 + 3.9057345

Rotational Inertia = 5.207646kgm²

Approximately = 5.21kgm²

b. The magnitude of the rod's angular momentum about the rotational axis is calculated as

Rotational Inertia about its axis × angular speed of the rod.

Angular speed of the rod is calculated as= (Angular velocity of the rod × 2π)/60

= (226×2π) /60

= 23.67 rad/s

Rotational Inertia = 5.21kgm²

The magnitude of the rod's angular momentum about the rotational axis

= 5.21kgm²× 23.67 rad/s

= 123.3207kgm²/s

Approximately = 123.32kgm²/s

7 0

3 years ago