An 8N force acts on a rectangular conductor 20cm long placed perpendicular to a magnetic field. Determine the magnetic field induction if the current in the conductor is 40A.

Hint: Here a current carrying wire or conductor when placed in a magnetic field experiences a force. If the direction of the magnetic field and the direction of the wire or conductor are in 90 degrees with each other then the force that will be acting on the conductor would be perpendicular to both the magnetic field and the current carrying conductor. It can be determined using Fleming’s Left Hand Rule. Here apply the formula $F = BiL\sin \theta $; where F = Force; B = Magnetic Field Induction; i = current; L = Length of the conductor.

Complete step by step solution:
Put the known value in the formula and find B.
$F = BiL\sin \theta $
Take B to LHS and put rest of the variables in RHS
$\Rightarrow$ $\dfrac{F}{{iL\sin \theta }} = B$;
Put the given value in the above equation and solve,
$\Rightarrow$ $\dfrac{8}{{40 \times 0.2 \times \sin 90}} = B$; ….(Here $L = 20cm = 0.2m$)
Do the necessary mathematical calculation and solve for “B”.
$\Rightarrow$ $B = \dfrac{8}{8}$
The final value of B is:
$\Rightarrow$ $B = 1$ $Tesla$
The magnetic field induction is $B = 1$ $Tesla$.

Note: Here make sure to apply the correct formula for force on current carrying wire placed in a magnetic field. Do not use the formula $F = qvb\sin \theta $. Here we have been given the value of length of the wire and current in the wire. We need to use a formula ($F = BiL\sin \theta $) that relates all the given variables together.