Question

7) A copper wire having resistance 0.01 ohm in each meter is used to wind a 400 turn
solenoid of radius 1 cm and length 20 cm. Find the e.m.f of a battery which when
connected across the solenoid will causes a magnetic field of 1x10-2 T near the center of
the solenoid.
[ANS: 1 V]

Answer 1

$\boxed{Magnetosatics} </strong></p><p></p><p><strong>$

Final Answer : 1 V

Steps:

1) We have,

Magnetic Field in centre of solenoid,

B = $1 * 10^{-2} T$

No. of turns per unit length,

$n =\frac{400}{0.20 m }= 2 *10^{3} \:units</strong></p><p></p><p><strong>$

Assuming there is no internal resistance in battery. By Ohm's Law :

$E = IR$

It is pretty obvious to say :

$R = 0.01 * 400* 2\pi *10^{-2} = 8\pi *10^{-2} \Omega</strong></p><p></p><p><strong>$

2) Magnetic Field at centre of Solenoid,

$B =\mu_o n I </strong></p><p></p><p><strong>\\ => 1*10^{-2}T = 4\pi *10^{-7}*2*10^3 * \frac{E}{R} </strong></p><p></p><p><strong>\\ </strong></p><p></p><p><strong>=> E = \frac{10^{-2}*8\pi *10^{-2}}{2*10^3*4\pi *10^{-7}} \\ </strong></p><p></p><p><strong>=> E = 1 V$

Hence, EMF of Battery is 1 V.