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
$\varphi =B(\pi r^2)$
$e=\frac{d\varphi }{dt}=\pi r^2(\frac{dB}{dt})$
(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 = $\varphi \frac{1}{\pi a^2}....(ii)$
where $l=2\pi r;m=\pi a^{2}l\sigma$
On putting foe value of in $a^2$ in eq. $(ii)$, we get
$i=\frac{m}{4\pi \rho \sigma }\frac{dB}{dt}$