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

Answer:

Current in wire will be equal to $i=2.376\times 10^9A$

Explanation:

We have given length of the wire l = 4.40 mm $=4.40\times 10^{-3}m$

Maximum force $F=0.575MN=0.575\times 10^6N$

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 $F=iBl\times 1=iBl$

So $0.575\times 10^6=i\times 0.0550\times 4.40\times 10^{-3}$

$i=2.376\times 10^9A$

So current in wire will be equal to $i=2.376\times 10^9A$