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Solution

The components of the velocity vector are u x = −5.0 , u y = −2.0 , and u z = 3.5 .

(a) The components of the magnetic field vector are B x = 7.2 , B y = −1.0 , and B z = −2.4 . Substituting them into [link] gives the scalar components of vector F = ζ u × B :

{ F x = ζ ( u y B z u z B y ) = ζ [ ( −2.0 ) ( −2.4 ) ( 3.5 ) ( −1.0 ) ] = 8.3 ζ F y = ζ ( u z B x u x B z ) = ζ [ ( 3.5 ) ( 7.2 ) ( −5.0 ) ( −2.4 ) ] = 13.2 ζ F z = ζ ( u x B y u y B x ) = ζ [ ( −5.0 ) ( −1.0 ) ( −2.0 ) ( 7.2 ) ] = 19.4 ζ .

Thus, the magnetic force is F = ζ ( 8.3 i ^ + 13.2 j ^ + 19.4 k ^ ) and its magnitude is

F = F x 2 + F y 2 + F z 2 = ζ ( 8.3 ) 2 + ( 13.2 ) 2 + ( 19.4 ) 2 = 24.9 ζ .

To compute angle θ , we may need to find the magnitude of the magnetic field vector,

B = B x 2 + B y 2 + B z 2 = ( 7.2 ) 2 + ( −1.0 ) 2 + ( −2.4 ) 2 = 7.6 ,

and the scalar product F · B :

F · B = F x B x + F y B y + F z B z = ( 8.3 ζ ) ( 7.2 ) + ( 13.2 ζ ) ( −1.0 ) + ( 19.4 ζ ) ( −2.4 ) = 0 .

Now, substituting into [link] gives angle θ :

cos θ = F · B F B = 0 ( 18.2 ζ ) ( 7.6 ) = 0 θ = 90 ° .

Hence, the magnetic force vector is perpendicular to the magnetic field vector. (We could have saved some time if we had computed the scalar product earlier.)

(b) Because vector B = 4.5 k ^ has only one component, we can perform the algebra quickly and find the vector product directly:

F = ζ u × B = ζ ( −5.0 i ^ 2.0 j ^ + 3.5 k ^ ) × ( 4.5 k ^ ) = ζ [ ( −5.0 ) ( 4.5 ) i ^ × k ^ + ( −2.0 ) ( 4.5 ) j ^ × k ^ + ( 3.5 ) ( 4.5 ) k ^ × k ^ ] = ζ [ −22.5 ( j ^ ) 9.0 ( + i ^ ) + 0 ] = ζ ( −9.0 i ^ + 22.5 j ^ ) .

The magnitude of the magnetic force is

F = F x 2 + F y 2 + F z 2 = ζ ( −9.0 ) 2 + ( 22.5 ) 2 + ( 0.0 ) 2 = 24.2 ζ .

Because the scalar product is

F · B = F x B x + F y B y + F z B z = ( −9.0 ζ ) ( 0 ) + ( 22.5 ζ ) ( 0 ) + ( 0 ) ( 4.5 ) = 0 ,

the magnetic force vector F is perpendicular to the magnetic field vector B .

Significance

Even without actually computing the scalar product, we can predict that the magnetic force vector must always be perpendicular to the magnetic field vector because of the way this vector is constructed. Namely, the magnetic force vector is the vector product F = ζ u × B and, by the definition of the vector product (see [link] ), vector F must be perpendicular to both vectors u and B .

Check Your Understanding Given two vectors A = i ^ + j ^ and B = 3 i ^ j ^ , find (a) A × B , (b) | A × B | , (c) the angle between A and B , and (d) the angle between A × B and vector C = i ^ + k ^ .

a. −2 k ^ , b. 2, c. 153.4 ° , d. 135 °

Got questions? Get instant answers now!

In conclusion to this section, we want to stress that “dot product” and “cross product” are entirely different mathematical objects that have different meanings. The dot product is a scalar; the cross product is a vector. Later chapters use the terms dot product and scalar product interchangeably. Similarly, the terms cross product and vector product are used interchangeably.

Summary

  • There are two kinds of multiplication for vectors. One kind of multiplication is the scalar product, also known as the dot product. The other kind of multiplication is the vector product, also known as the cross product. The scalar product of vectors is a number (scalar). The vector product of vectors is a vector.
  • Both kinds of multiplication have the distributive property, but only the scalar product has the commutative property. The vector product has the anticommutative property, which means that when we change the order in which two vectors are multiplied, the result acquires a minus sign.
  • The scalar product of two vectors is obtained by multiplying their magnitudes with the cosine of the angle between them. The scalar product of orthogonal vectors vanishes; the scalar product of antiparallel vectors is negative.
  • The vector product of two vectors is a vector perpendicular to both of them. Its magnitude is obtained by multiplying their magnitudes by the sine of the angle between them. The direction of the vector product can be determined by the corkscrew right-hand rule. The vector product of two either parallel or antiparallel vectors vanishes. The magnitude of the vector product is largest for orthogonal vectors.
  • The scalar product of vectors is used to find angles between vectors and in the definitions of derived scalar physical quantities such as work or energy.
  • The cross product of vectors is used in definitions of derived vector physical quantities such as torque or magnetic force, and in describing rotations.
Practice Key Terms 6

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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