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3 years ago

In which direction does magnetic field in the center of the coil point?

Physics

2 answers:

Marina86 [1]3 years ago

8 0

Answer:

Right wards

Explanation:

The direction of magnetic field in a loop is given by Maxwell's right hand thumb rule.

According to this rule, if we hold a straight current carrying conductor in our right hand such that the thumb indicates the direction of current, then the grasping of fingers indicates the direction of magnetic field around the conductor and vice versa.

So, here the direction of magnetic filed is in rightwards direction.

denis-greek [22]3 years ago

7 0

Answer:

Magnetic Field from a Loop. The direction of the field is given by another right-hand rule. Curl the fingers on your right hand the way the current goes. Stick your thumb out and it points in the direction of the magnetic field inside the loop.

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kifflom [539]

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1) A pendulum is configured to have a period of 2 seconds.

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3 years ago

in a cricket match there are 5000 spectators counted 10 by 10 the number of significant figure in the measurement will be

BabaBlast [244]

Answer:

$\boxed{3 \ Significant \ figures}$

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3 years ago

Mass, volume and density are all properties of

snow_tiger [21]

Answer:

Properties of matter

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7 0

3 years ago

Read 2 more answers

Two identical conducting spheres, having charges of opposite sign, attract each otherwith a force of 0.108 n when separated by 5

lisov135 [29]

Let's start from the final situation. After the two spheres are connected with the conducting wire, the total charge distributes equally between the two spheres (because they are identical). We can call the charge on each sphere $Q/2$ , with Q being the total charge.
The electrostatic force in this situation is 0.0360 N, so we can write
$F=k \frac{( \frac{Q}{2} )^2}{r^2}$
where k is the Coulomb's constant and $r=50.0 cm=0.50 m$ is the separation between the two spheres. Using F=0.0360 N, we can find the value of Q, the total charge shared between the two spheres:
$Q= \sqrt{ \frac{4Fr^2}{k} } = \sqrt{ \frac{4(0.0360 N)(0.50 m)^2}{8.99 \cdot 10^9 Nm^2C^{-2}} }=2.0 \cdot 10^{-6}C$

Now let's go back to the initial situation, before the conducting wire was attached; in this situation, the two spheres have a charge of $q_1$ and $q_2$ , whose sum is Q:
$Q=q_1 + q_2$
The electrostatic force between the two spheres in the initial situation is:
$F=k \frac{q_1 q_2}{r^2}$
And since we know F=0.108 N, we find
$q_1 q_2 = \frac{Fr^2}{k} = \frac{(0.108 N)(0.50 m)^2}{8.99 \cdot 10^9 N m^2 C^{-2}}=3.0 \cdot 10^{-12} C$
But the problem tells us that the two spheres have charges of opposite sign, so we must put a negative sign:
$-3.0 \cdot 10^{-12} C = q_1 q_2$

So now we have basically a system of 2 equations:
$2.0 \cdot 10^{-6} C = Q = q_1 + q_2$
$-3.0 \cdot 10^{-12} C = q_1 q_2$
If we solve it, we find the initial charge on the two spheres:
$q_1 = -1 \cdot 10^{-6}C$
$q_2 = +3 \cdot 10^{-6 } C$

6 0

3 years ago