Ask question

Ask question

All categories

nataly862011 [7]

3 years ago

13

1 What happens to the direction of the force on a current-carrying wire if the

1 answer:

sertanlavr [38]3 years ago

5 0

Answer:

<h2>If the forefinger points in the direction of the magnetic field and the middle finger in the direction of the current, then, the thumb points in the direction of the force exerted on the conductor.So, if the current is reversed in the conductor placed in a magnetic field the direction of force is also reversed..</h2>

You might be interested in

Which of these is a benefit of social networking ?

Alisiya [41]

Answer:

Staying connected to friends

Explanation:

hope this helps

6 0

2 years ago

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is th

katrin [286]

Complete Question

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is the nail compressed if it is 2.50 mm in diameter and 6.00-cm long.What Average force is excreted on the Nail

Answer:

$F=2*10^{4}N$

Explanation:

From the question we are told that:

Mass $m=0.500kg$

Initial Velocity $V=15.0m/s$

Distance $x=2.80mm=>0.00280m$

Diameter $d=2.50mm=>0.00250m$

Length $l=6.00cm=>0.6m$

Generally the equation for Force is mathematically given by

$F=\frac{mv^2}{2d}$

$F=\frac{0.500*15^2}{2.80*10^{-3}}$

$F=2*10^{4}N$

6 0

2 years ago

Question 2 of 10

Mekhanik [1.2K]

the awnser to ur question is D

6 0

3 years ago

How can juggling improve your hand eye coordination?

Ne4ueva [31]

Answer:

you can predict where the juggling ball is going to land and the move you hand to catch it

Explanation:

5 0

3 years ago

Read 2 more answers

A certain elastic conducting material is stretched into a circular loop of 1.6 cm radius. It is placed with its plane perpendicu

inysia [295]

Answer:

Induced emf in the loop is 0.0603 volts.

Explanation:

It is given that,

Radius of the circular loop, r = 1.6 cm = 0.016 m

Magnetic field, B = 0.8 T

When released, the radius of the loop starts to shrink at an instantaneous rate of 75.0 cm/s, $\dfrac{dr}{dt}=75\ cm/s=0.75\ m/s$

We need to find the magnitude of induced emf at that instant. Induced emf is given by :

$\epsilon=\dfrac{-d\phi}{dt}$

Where

$\phi$ is the magnetic flux, $\phi=B\times A$

$\epsilon=\dfrac{-d(BA)}{dt}$ , A is the area of cross section

$\epsilon=-\dfrac{-d(B(\pi r^2))}{dt}$

$\epsilon=-2\pi r B(\dfrac{dr}{dt})$

$\epsilon=-2\pi \times 0.016\times 0.8 \times 0.75\ m/s$

$\epsilon=0.0603\ V$

So, the induced emf in the loop is 0.0603 volts. Hence, this is the required solution.

3 0

3 years ago