# A solenoid with a cross-sectional area of 1.75 x 10^{-3} m^2 is 0.780 m long and has 450 turns...

## Question:

A solenoid with a cross-sectional area of {eq}\rm 1.75 \times 10^{-3} \ m^2 {/eq} is 0.780 m long and has 450 turns per meter. Find the magnitude of the induced emf in this solenoid if the current in it is increased from 0 to 1.65 A in 50.5 ms.

## Solenoid and the Induced Emf:

The long coil that is wrapped in no. of turns is known as Solenoid. The magnetic field inside the solenoid is directly proportional to the no. of turns, current in the solenoid.

Given data

Area of the cross-section of the solenoid {eq}(A) = 1.75 \times 10^{-3} \ m^{2} {/eq}

No. of turns per unit length {eq}(n) = 450 \ turns/meter {/eq}

Now, the magnetic field inside the solenoid is given by

{eq}B = \mu_{o}nI {/eq}

where

• I is the current in the solenoid

Now, the induced emf is given by

{eq}E = \dfrac{(B_{2} -B_{1})A }{\Delta t} \\ E = \dfrac{(A)\mu_{o}n (I_{2} - I_{1})}{\Delta t} \\ E = \dfrac{(1.75 \times 10^{-3}) (4\pi \times 10^{-7})(450)(1.65- 0)}{50.5 \times 10^{-3} \ s} \\ E = 3.23 \times 10^{-5} \ Volt {/eq}

where

• {eq}I_{2} = 1.65 \ A {/eq} is the final current in the solenoid
• {eq}(I_{1}) = 0 \ A {/eq} is the initial current in the solenoid
• {eq}(\Delta t ) = 50.5 \times 10^{-3} \ s {/eq} is the time of changing the value of current 