Ask question

Ask question

All categories

2 months ago

6

To practice Problem-Solving Strategy 29.1: Faraday's Law. A metal detector uses a changing magnetic field to detect metallic obj

ects. Suppose a metal detector that generates a uniform magnetic field perpendicular to its surface is held stationary at an angle of 15.0∘∘ to the ground, while just below the surface there lies a silver bracelet consisting of 6 circular loops of radius 5.00 cmcm with the plane of the loops parallel to the ground. If the magnetic field increases at a constant rate of 0.0250 T/sT/s, what is the induced emf EEEMF? Take the magnetic flux through an area to be positive when B⃗ B→B_vec crosses the area from top to bottom.

1 answer:

kicyunya [3.2K]2 months ago

4 0

Response:

$1.138\times 10^{-3}V$

Details:

Utilizing Faraday's Newmann Lenz law enables the assessment of the induced emf within the loop:

$\epsilon=\frac{d\phi}{dt}$

where:

$d\Phi-$ represents the change in magnetic flux

$dt-$ symbolizes the change over time.

#The magnetic flux linked to the coil can be represented as:

$\Phi=NBA \ Cos \theta$

Where:

N represents the number of loops.

A denotes the area for each loop ( $A=\pi r^2=\pi \times 5^2=78.5398$ ).

B indicates the strength of the magnetic field.

$\theta=15\textdegree$ represents the angle between the magnetic field direction and the normal to the loop's area.

$\epsilon=-\frac{d(78.5398\times 10^{-3}NB \ Cos \theta)}{dt}\\\\=-(78.5398\times 10^{-3}N\ Cos \theta)}{\frac{dB}{dt}$

$\frac{dB}{dt}-$ =0.0250T/s is indicated as the rate of magnetic field increase.

#Plugging in values into the emf equation:

$=-(78.5398\times 10^{-3} m^2 \times 6\ Cos 15 \textdegree)\times 0.0250T/s\\\\=1.138\times 10^{-3}V$

Thus, the induced emf is $1.138\times 10^{-3}V$

You might be interested in

2.27 A gas is compressed from V1= 0.3 m3, p1=1 bar to V2= 0.1 m3, p2 =3 bar. The pressure and

Yuliya22 [3333]

The work done is -40 kJ and also equals 80 kJ. Explanation: Work is represented by the area under the pressure versus volume graph.

5 0

1 month ago

A scale used to weigh fish consists of a spring hung from a support. The spring's equilibrium length is 10.0 cm. When a 4.0 kg f

ValentinkaMS [3465]

A) 1153 N/m

We determine the spring constant using Hooke's law:

$F=kx$

where

F represents the force exerted on the spring

k denotes the spring constant

x signifies the spring's displacement

In this case, a fish with mass m = 4.0 kg is hanging from the spring, and hence, the force applied is the fish's weight:

$F=mg=(4.0 kg)(9.8 m/s^2)=39.2 N$

and the spring's displacement is:

$x = 13.4 cm - 10.0 cm = 3.4 cm = 0.034 m$

thus the spring constant amounts to

$k=\frac{F}{x}=\frac{39.2 N}{0.034 m}=1153 N/m$

B) 16.8 cm

For the second situation, a fish of mass

m = 8.0 kg

is attached to the spring. Hence, the force applied to the spring is

$F=mg=(8.0 kg)(9.8 m/s^2)=78.4 N$

Thus, we can ascertain the spring's displacement:

$x=\frac{F}{k}=\frac{78.4 N}{1153 N/m}=0.068 m = 6.8 cm$

Considering the spring's equilibrium length of

$x_0 = 10.0 cm$

the new length of the spring becomes

$x' = 10.0 cm + 6.8 cm = 16.8 cm$

5 0

1 month ago

Three equal negative point charges are placed at three of the corners of a square of side d. What is the magnitude of the net el

Ostrovityanka [3204]

<span>this might be useful
Regarding the field, the two charges placed opposite cancel each other out!
Therefore, E = kQ / d² = k * Q / (d/√2)² = 2*k*Q / d² ◄
given k = 8.99×10^9 N·m²/C²,
E = 1.789×10¹⁰ N·m²/C² * Q / d² </span>

8 0

2 months ago

Read 2 more answers

What is the energy equivalent of an object with a mass of 2.5 kg? 5.5 × 108 J 7.5 × 108 J 3.6 × 1016 J 2.25 × 1017 J

Maru [3345]

Utilizing Einstein's equation that signifies the relationship between mass and energy:
$E=mc^2$
where
E stands for energy
m represents the object's rest mass
and c denotes the speed of light

By applying this formula and using m=2.5 kg, we can compute the object's equivalent energy:
$E=mc^2 = (2.5 kg)(3 \cdot 10^8 m/s)^2=2.25 \cdot 10^{17} J$

6 0

1 month ago

Read 2 more answers

A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first bl

Softa [3030]

According to Newton's second law, provided F is directed horizontally,

• the net horizontal force acting on the larger block is

F - µmg = 3mA

where µmg represents the friction experienced by the larger block from contact with the smaller block, µ is the static friction coefficient between both blocks, and A indicates the acceleration of the block;

• the net vertical force on the larger block is

4mg - 3mg - mg = 0

where 4mg denotes the magnitude of the normal force exerted by the surface on the combined mass of both blocks, 3mg corresponds to the weight of the larger block, and mg indicates the weight of the smaller block;

• the net horizontal force acting on the smaller block is

µmg = ma

where µmg again signifies the friction between both blocks; however, it is important to note that this force aligns in the same direction as F. It is the sole force influencing the smaller block in the horizontal direction, thus (b) static friction causes the smaller block's acceleration;

• the net vertical force on the smaller block is

mg - mg = 0

where mg represents the force of both the normal force from the larger block pushing up against the smaller one, and the weight of the smaller block.

(You should be able to create your own free-body diagrams based on the forces discussed.)

(c) Solve the equations stated above to find A and a:

A = (F - µmg) / (3m)

a = µg

5 0

1 month ago