2. physical quantities which do not depend on any physical quantities for their measurements are known as

A circular loop of radius r carries a current i. at what distance along the axis of the loop is the magnetic field one-half its

lana [24]

At r = 0.766 R the magnetic field intensity will be half of its value at the center of the current carrying loop.

We have a circular loop of radius ' r ' carrying current ' i '.

We have to find at what distance along the axis of the loop is the magnetic field one-half its value at the center of the loop.

<h3>What is the formula to calculate the Magnetic field intensity due to a current carrying circular loop at a point on its axis?</h3>

The formula to calculate the magnetic field intensity due to a current carrying ( i ) circular loop of radius ' R ' at a distance ' x ' on its axis is given by -

$B(x) = \frac{\mu_{o} iR^{2} }{2(x^{2} +R^{2})^{\frac{3}{2} } }$

Now, for magnetic field intensity at the center of the loop can calculated by putting x = 0 in the above equation. On solving, we get -

$B(0) = \frac{\mu_{o} i}{2R}$

Let us assume that the distance at which the magnetic field intensity is one-half its value at the center of the loop be ' r '. Then -

$\frac{\mu_{o} iR^{2} }{2(r^{2} +R^{2})^{\frac{3}{2} } } = \frac{1}{2} \frac{\mu_{o}i }{2R}$

$2R^{3} = (r^{2} +R^{2} )^{\frac{3}{2} }$

$4R^{6} = (r^{2} +R^{2} )^{3}$

$r^{2} =0.587R^{2}$

r = 0.766R

Hence, at r = 0.766 R - the magnetic field intensity will be half of its value at the center of the current carrying loop.

To solve more questions on magnetic field intensity, visit the link below-

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6 0

2 years ago

Batteries supply DC, direct current, which does not change in direction over time, though eventually they get exhausted.

SashulF [63]

My Anonymuos friend, I could not agree more with your statement.
Truer words have rarely if ever been expressed.

6 0

3 years ago

If you push twice as hard against a stationary brick wall, the amount of work you do

Alecsey [184]

Answer:

Zero

Explanation:

The work done on an object is given by:

$W=Fd cos \theta$

where

F is the force applied on the object

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement

In this problem, you are pushing again a stationary wall: this means that the walls does not move. As a result, the displacement is zero: d=0. Therefore, the work done is also zero: W=0.

8 0

3 years ago

a block measuring 20cm by 10cm by 5cm rests on a flat surface. The block has a weight of 3N. Determine the maximum pressure it e

Lady_Fox [76]

Max preassure = force / min area
= 3N / 0.1 x 0.05
= 600N/m(squared)
Copy off of the picture below itll help better, its what someone sent me when i asked this question

4 0

4 years ago

A 2.00-kg object traveling east at 20.0 m/s collides with a 3.00-kg object traveling west at 10.0 m/s.

ELEN [110]

Momentum = mass x velocity

Before collision
Momentum 1 = 2 kg x 20 m /s = 40 kg x m/s
Momentum 2 = 3 kg x -10m/s = -30 kg x m/s

After collision
Momentum 1 = 2 kg x -5 m/s = -10 m/s
Momentum 2 = 3 kg x V2 = 3V2

Total momentum before = total momentum after
40 + -30 = -10 + 3V2
V2 = <span>6.67 m/s

Total kinetic energy before
</span><span>= (1/2) [ 2 kg * 20 m/s * 2 + 3 kg * ( -10 m/s) *2 ]
= 550 J
</span>
<span>Total kinetic energy after
</span>= (1/2) [ 2 kg * ( - 5 m/s) * 2 + 3 kg * 6.67 m/s *2 ]
= 91.73 J

Total kinetic energy lost during collision
=<span>550 J - 91.73 J
= 458.27 J</span>

8 0

3 years ago

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2. physical quantities which do not depend on any physical quantities for their measurements are known as​

2. physical quantities which do not depend on any physical quantities for their measurements are known as