Understanding the Red-Boxed Formula

[gsan]

Refer to the attachment below,

anyone can explain the formula highlighted by red box? how the | dl x R | become dl only?

 

[Doc Al]

Note that $$\hat{R}$$ is a unit vector that is perpendicular to $$\vec{d\ell}$$.

 

[gsan]

Note that $$\hat{R}$$ is a unit vector that is perpendicular to $$\vec{d\ell}$$.

yes, I'm understand that dl is perpendicular to unit vector of R, but how the magnitude of them become dl only?

 

[Doc Al]

yes, I'm understand that dl is perpendicular to unit vector of R, but how the magnitude of them become dl only?

The magnitude of any cross product is given by
$$\vec{A}\times\vec{B} = AB\sin\theta$$

Applying this to your question, B is a unit vector and theta = 90, so
$$AB\sin\theta = A$$

Make sense?

 

[gsan]

The magnitude of any cross product is given by
$$\vec{A}\times\vec{B} = AB\sin\theta$$

Applying this to your question, B is a unit vector and theta = 90, so
$$AB\sin\theta = A$$

Make sense?

I not really understand about the 2nd formula, could explain more?

 

[Doc Al]

I not really understand about the 2nd formula, could explain more?

What aspect do you not understand? A = dl; B = 1; theta = 90.

Have you worked with vector cross products before?

 

[gsan]

What aspect do you not understand? A = dl; B = 1; theta = 90.

Have you worked with vector cross products before?

why the B=1 ?

yes, I know that
$$\vec{A}\times\vec{B} = AB\sin\theta$$

 

[Doc Al]

why the B=1 ?

It represents the magnitude of the unit vector $$\hat{R}$$, which is 1.

Perhaps you are confusing $$\vec{R}$$, which has magnitude of R, with $$\hat{R}$$, which has magnitude of 1?

 

[gsan]

It represents the magnitude of the unit vector $$\hat{R}$$, which is 1.

Perhaps you are confusing $$\vec{R}$$, which has magnitude of R, with $$\hat{R}$$, which has magnitude of 1?

how do u know the $$\hat{R}$$ has magnitude of 1 ?

 

[Doc Al]

how do u know the $$\hat{R}$$ has magnitude of 1 ?

It's a unit vector! The purpose of a unit vector is to define a direction; unit vectors--by definition--have magnitude of 1.

The little "hat" symbol ^ on top of the vector tells you that it's a unit vector. Its direction is the direction of the vector R, but its magnitude is 1.

 

[gsan]

ok, I understand why unit vector of R has magnitude of 1.

now my another question is, why we can cross product of a vector with a unit vector?
eg. vector of dl x unit vector of R

 

[Doc Al]

now my another question is, why we can cross product of a vector with a unit vector?

Why not? A unit vector is a perfectly good vector, just like any other.

 

[gsan]

let say vector of dl = hat{x} dl and vector of R = hat{x}3 + hat{y}2 + hat{z} 3

the cross product of vector of dl x vector of R should be

|hat{x} hat{y} hat{z}|
| dl 0 0 |
| 3 2 3 |

or

|hat{x} hat{y} hat{z}|
| 1 0 0 |
| 3 2 3 |

 

[Doc Al]

True. But realize that here you need the cross product of vector dl and unit vector R hat, not vector R. The magnitude of R hat must equal 1.

Given your definition of vector R, what would be unit vector R hat? (Hint: Find the magnitude of vector R.)

 

[gsan]

True. But realize that here you need the cross product of vector dl and unit vector R hat, not vector R. The magnitude of R hat must equal 1.

Given your definition of vector R, what would be unit vector R hat? (Hint: Find the magnitude of vector R.)

vector R = (unit vector R)(magnitude of R)

and now my question is

let say vector of dl = hat{x} dx and vector of R = hat{x}3 + hat{y}2 + hat{z} 3

the cross product of vector of dl and vector of R should be

|hat{x} hat{y} hat{z}|
| dx 0 0 |
| 3 2 3 |

or

|hat{x} hat{y} hat{z}|
| 1 0 0 |
| 3 2 3 |

which 1 should be correct? 1st or the 2nd?

 

[Doc Al]

and now my question is

the cross product of vector of dl and vector of R should be

|hat{x} hat{y} hat{z}|
| dl 0 0 |
| 3 2 3 |

or

|hat{x} hat{y} hat{z}|
| 1 0 0 |
| 3 2 3 |

which 1 should be correct? 1st or the 2nd?

Assuming that the vector dl is purely in the x direction (for some reason), then version 1 is correct. Version 2 treats dl as a unit vector, which it is not.

In general, dl will be in some arbitrary direction.

 

[gsan]

Assuming that the vector dl is purely in the x direction (for some reason), then version 1 is correct. Version 2 treats dl as a unit vector, which it is not.

In general, dl will be in some arbitrary direction.

sorry, my mistake, let say the cable carry current lies along the x-axis,

so vector of dl = hat{x} dx and vector of R = hat{x}3 + hat{y}2 + hat{z} 3

the cross product of vector of dl and vector of R

|hat{x} hat{y} hat{z}|
| 1 0 0 |
| 3 2 3 |

[ - hat{y}3 + hat{z}2 ]dx

is my working correct?

 

[Doc Al]

Yes, but the vector hat{x} dx should have components dx, 0, 0, not 1, 0, 0.

 

[gsan]

An infinite current filament carries a current of 3A and lies along the x-axis. Using Biot-Savart Law, find the magnectic field intensity in cartesian coordinates at a point P(-1,3,2).

dH = I (vet)dl x hat{R} / 4piR^2

let say substitude hat{R} with (vet)R / R

then dH = I (vet)dl x (vet)R / 4piR^3

R= hat{x} (-1-x) + hat{y} 3 + hat{z}2

is my working correct? and what should i do for the next step?

 

[Doc Al]

Unless you are being asked to integrate to find the field at that point, you are doing this the hard way. What's the field from an infinitely long current-carrying wire?

 

[gsan]

ya, I'm asked to integrate to find the field at that point