Question

A horizontal magnetic field B is produced across a narrow gap between the two square iron pole pieces. A closed square loop of side a, mass m and resistance R is allowed to fall with the top of the loop in the field. The loop attains a terminal velocity equal to :

• A. $\frac{Rmg}{B^2{a}^2}$
• B. $\frac{{\mu }_0}{4\pi }\frac{Rmg}{B{a}^2}$
• C. $\frac{mgB}{R{a}^2}$
• D. $\frac{{\mu }_0}{4\pi }\frac{RmB}{B{a}^2}$

Answer 1

The correct option is A $\frac{Rmg}{B^2{a}^2}$

Force on wire $i\overrightarrow{l}\times \overrightarrow{B}$
area of$\varphi =xa$
$\frac{dx}{dt}=-V$
$\varphi =Bax$
$e=\frac{-d\varphi }{dt}=BaV\Rightarrow i=\frac{BaV}{R}$
the magnetic forces on AD and BC are cancelled by each other. Magnetic field applies upward force on CD wires
$F=iBa=\frac{B^2{a}^{2}V}{R}$
which is equal to mg
$\frac{B^2{a}^{2}V}{R}=mg$
$V=\frac{Rmg}{B^2{a}^2}$