Question

A 0.5 long solenoid of 10 turns/cm has area of cross section 1 cm2 cal voltage induced across its ends if the current in the solenoid changes from 1ampere to 2 ampere in o. 1 sec

Answer 1

Answer: The voltage induced across the ends is $\epsilon = 6.25\times10^{-6} volt$

Explanation:

Given that,

Length l = 0.5 m

Area of cross section $A = 1\times10^{-4}m^2$

Changing current $\Delta I=2-1 = 1\ A$

Changing time $\Delta t= 0.1 s$

Number of turns $N = 50\times10= 500$

Using formula of magnetic field

$\Delta B=\mu_{0} n\Delta I$

$\Delta B= 4\times3.14\times10^{-7}\times10\times1$

$\Delta B= 1.25\times10^{-5} T$

The voltage induced is

$\epsilon= N\dfrac{d\phi}{dt}$

$\epsilon= N\dfrac{\Delta B\cdot A}{\Delta t}$

$\epsilon= 500\times\dfrac{1.25\times10^{-5}\times 1\times10^{-4}}{0.1}$

$\epsilon = 6.25\times10^{-6} volt$

Hence, The voltage induced across the ends is $\epsilon = 6.25\times10^{-6} volt$.

Answer 2

Answer:

here is your answer

Explanation: