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zlopas [31]
3 years ago
5

A rectangular loop with dimensions 4.20 cm by 9.50 cm carries current I. The current in the loop produces a magnetic field at th

e center of the loop that has magnitude 3.10×10−5 T and direction away from you as you view the plane of the loop. Part A What is the magnitude of the current in the loop?
Physics
1 answer:
tiny-mole [99]3 years ago
8 0

Answer:

I=1.48 A

Explanation:

Given that

B=3.1 x 10⁻5 T

b= 4.2 cm

l= 9.5 cm

The relationship for magnetic field  and current given as

B=\dfrac{2\mu _oI}{\pi}D

Where

D=\dfrac{\sqrt{l^2+b^2}}{lb}

By putting the values

D=\dfrac{\sqrt{l^2+b^2}}{lb}

D=\dfrac{\sqrt{0.042^2+0.095^2}}{0.042\times 0.095}

D=26.03 m⁻¹

B=\dfrac{2\mu _oI}{\pi}D

3.1\times 10^{-5}=\dfrac{2\times 4\times \pi \times 10^{-7} I}{\pi}\times 26.03

I=\dfrac{3.1\times 10^{-5}}{{2\times 4\times 10^{-7} }\times 26.03}

I=1.48 A

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A stockroom worker pushes a box with mass 11.2 kg on a horizontal surface with constant speed of 3.50m/s. The coefficient of kin
stellarik [79]

Answer:

3.125 m

Explanation:

We are given that

Mass of box=m=11.2 kg

Speed of box=u=3.5m/s

Coefficient of kinetic friction=\mu_k=0.2

Final velocity,v=0

a.We have to find the horizontal force applied by worker to maintain the motion.

According to question

Horizontal force=F=f=\mu_kmg

g=9.8m/s^2

Substitute the values

Horizontal force=F=0.2\times 11.2\times 9.8=21.95 N

b.According to work-energy theorem

W=\frac{1}{2}mv^2-\frac{1}{2}mu^2

-\mu mg s=\frac{1}{2}(11.2)(0)^2-\frac{1}[2}(11.20)(3.5)^2

-\mu mg s=-\frac{1}{2}(11.2)(3.5)^2

0.2\times (11.2)\times 9.8\times s=\frac{1}{2}(11.2)(3.5)^2

s=\frac{11.2\times (3.5)^2}{2\times 0.2\times 11.2\times 9.8}

s=3.125 m

Hence, the box slide before coming to rest=3.125 m

4 0
4 years ago
Can you help me answer this?
Pavel [41]
The answer is D. If you aren't consistent with your drop positions, then your data may be invalid. To be frank: it basically screws over the experiment.
5 0
3 years ago
What is the angular speed of (a) the second hand, (b) the minute hand, and (c) the hour hand of a smoothly running analog watch?
Agata [3.3K]
A)We know the formula of the angular speed is                 ω = 2π / TWhere T is the time period.When second hand completes one revolution then the time taken is 60s.So T = 60sThen the angular speed of the second hand is          ω= 2π / (60s)             = 0.1047 rad/sb)When the minute hand completes one revolution the time taken is             T = 1 hr                = 3600sThen the angular speed of the minute hand is         ω =(2π) / (3600s) = 0.001745 rad/sc)When the hour hand completes one revolution then the timeperiod is          T = 12hrs             = (12)(3600)sThen the angular speed of the hour hand is         ω =(2π) / [(12)(3600)s] = 1.45444 x 10^-4 rad/s
8 0
4 years ago
Motions need an unbalanced net force to maintain.<br> True or False?
Musya8 [376]

Answer:

The answer is False

3 0
3 years ago
Read 2 more answers
Một dây dẫn đặt trong không khí có dòng điện I = 12A chạy qua, được gấp thành hình vuông cạnh a = 10cm. Xác định vectơ cường độ
disa [49]

Answer:

The net magnetic field ta the center of square is1.36\times10^{-4} T.

Explanation:

Current, I = 12 A , side ,a = 10 cm =  0.1 m

Let the magnetic field due to the one side is B.

The magnetic field is given by

B = \frac{\mu o}{4\pi}\times \frac{I}{r}\times \left (Sin A +Sin B  \right )\\\\B = 10^{-7}\times \frac{12}{0.05}\times \left ( sin 45 +  sin 45  \right )\\\\B = 3.4\times 10^{-5} T

Net magnetic field at the center of the square is

B' = 4 B

B'= 4\times 3.4\times 10^{-5}\\\\B' = 1.36\times10^{-4} T

4 0
3 years ago
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