Answer:
Current in wire will be equal to [tex]i=2.376\times 10^9A[/tex]
Explanation:
We have given length of the wire l = 4.40 mm [tex]=4.40\times 10^{-3}m[/tex]
Maximum force [tex]F=0.575MN=0.575\times 10^6N[/tex]
Magnetic field B = 0.0550 T
We know that force on wire is given by [tex]F=iBlsin\Theta[/tex]
For maximum force value of [tex]sin\Theta[/tex] will be maximum which is equal to 1
So force [tex]F=iBl\times 1=iBl[/tex]
So [tex]0.575\times 10^6=i\times 0.0550\times 4.40\times 10^{-3}[/tex]
[tex]i=2.376\times 10^9A[/tex]
So current in wire will be equal to [tex]i=2.376\times 10^9A[/tex]
