Question

The force on a wire carrying 7.35 A is a maximum of 2.33 N when placed between the pole faces of a magnet.

1) If the pole faces are 62.5 cm in diameter, what is the approximate strength of the magnetic field?

Answer 1

The expression to be used to find the force acting on a current carrying conductor kept in a magnetic field is given as F = B i l sin @.
B- magnetic field induction in tesla
i- the current in ampere
l - the length of the wire subjected to magnetic field
@ - the angle between the magnetic field vector and current element vector.
In your problem, @ is 90 as the force is said to be maximum.
For l, we have to take it to be 0.625 m assuming that the magnetic lines come out all along from the cross sectional face of poles.
Hence plugging and solving for B = 2.33/(7.35*0.625*sin 90 )
we get, B = 0.5072 tesla
Approximate magnetic field induction = 0.5072 tesla