Question

How much current is flowing in a wire 4.40 mm long if the maximum force on it is 0.575 NN when placed in a uniform 0.0550-TT field?

Answer 1

Current in wire will be equal to
`i=2.376* 10^9A`

Step-by-step explanation:

We have given length of the wire l = 4.40 mm
`=4.40* 10^(-3)m`

Maximum force
`F=0.575MN=0.575* 10^6N`

Magnetic field B = 0.0550 T

We know that force on wire is given by
`F=iBlsin\Theta`

For maximum force value of
`sin\Theta` will be maximum which is equal to 1

So force
`F=iBl* 1=iBl`

So
`0.575* 10^6=i* 0.0550* 4.40* 10^(-3)`

`i=2.376* 10^9A`

So current in wire will be equal to
`i=2.376* 10^9A`