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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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\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}.

At t = 20\; {\rm s}, the truck would have travelled a distance of x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}.

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Hence, the car would catch up with the truck at t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}.

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