Question

A magnetic field induction is changing in magnitude in a region at a constant rate dB/ dt A given mass m of copper drawn into a wire,and formed into a loop is placed perpendicular to the field. If the values of specific resistance and density of copper are $\rho$ and $\sigma$ respectively, then the current in , the loop is given by :

• A. $\frac{4\pi m}{\rho \sigma}\frac{dB}{dt}$
• B. $\frac{ m}{4\pi \rho \sigma}\frac{dB}{dt}$
• C. $\frac{ m}{\rho \sigma}\frac{dB}{dt}$
• D. $\frac{2\pi m}{\rho \sigma}\frac{dB}{dt}$

Answer 1

Answer: (B)

Let r is the radius of the loop. The magnetic flux, at a time t, through the loop is given by
$\phi = B (\pi r^2)$
$e = \frac{d \phi}{dt} = \pi r^2 ( \frac{dB}{dt})$
(the numerical valu of induced emf) , Induced current in the loop is given by ,br> $i = \frac{e}{R} = \frac{\pi r^2}{R} \frac{dB}{dt} \, ....(i)$
where R is the resistance of the wire let a is the corss section area and 1 the length of the wire.
Then R = $\phi \frac{1}{\pi a^2} \, ....(ii)$
where $l = 2 \pi r ; m = \pi a^2 l \sigma$
On putting foe value of in $a^2$ in eq. $(ii)$, we get
$i = \frac{m}{4 \pi \rho \sigma} \frac{dB}{dt}$