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ruslelena [56]
2 years ago
13

A conducting coil lies flat on a level tabletop in a region where the magnetic field points straight up. The magnetic field sudd

enly grows weaker. When viewed from above, what is the direction of the induced current in this coil as the field decreases?
a) clockwise
b) counterclockwise
c) clockwise initially, then counterclockwise before stopping
d) there is no induced current in this coil because the field is gone
Physics
1 answer:
Andrej [43]2 years ago
6 0

Answer:

clockwise

Explanation:

We shall apply Lenz's law to find the direction of induced current in case magnetic field changes. According to this law , direction of current induced is such that it will try to oppose the cause which produces the induced current .

In the present case, downward magnetic field decreases so current will be such that downward magnetic field increases . It can happen when clockwise current is induced. So clockwise current will be induced.

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Explanation:

1)

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It can be calculated using the following suvat equation for angular motion:

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In this problem we have:

\theta=90^{\circ} = \frac{\pi}{2}rad is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

Solving for \alpha, we find:

\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2

2)

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

\omega_f = \omega_i + \alpha t

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\omega_i is the initial angular velocity

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In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

Therefore, the final angular speed is:

\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s

3)

The tangential acceleration is related to the angular acceleration by the following formula:

a_t = \alpha r

where

a_t is the tangential acceleration

\alpha is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2

4)

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

v=u+at

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Here we have:

a=27900 m/s^2 (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

v=0+(27900)(0.0013)=36.3 m/s

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