Question

If A and B are irrotational, prove that A × B is Solenoidal that is div (A × B) = 0

Answer 1

By problem ∇ × A = 0 and ∇ × B = 0, it follows that

B .(∇ × A) = 0 

A. (∇ × B) = 0 

Subtracting, B .(∇ × A) − A.(∇ × B) = 0 

Now ∇ .(A × B) = B .(∇ × A) − A.(∇ × B) 

Therefore ∇ .(A × B) = 0, so that (A × B) is solenoidal

Answer 2

Proved below.

Step-by-step explanation:

Given:

Here, a and b are irrotational.  

To prove:

a b is solenoidal.

Proof:

By problem  ∇ x A = 0 and ∇ x B=0 , it is as follows

B.(∇ x A)=0                              [1]

A.(∇ x B)=0                               [2]

Subtracting Eq (2) from (1), we get

 B.∇(∇ x A) - A. (∇ x B)=0     [3]

Now,  

 ∇.(A x B)=B.(∇ x A)-A.(∇ x B)=0

Therefore,  

 ∇.(A x B)=0                 [from (3)]

so that (A x B) is a solenoid.

Hence proved.

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