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Question: A 0.5m long solenoid of 10turns/cm has an area of cross section \(1c{m^2}\). Calculate the voltage i...

A 0.5m long solenoid of 10turns/cm has an area of cross section 1cm21c{m^2}. Calculate the voltage induced across its ends if the current in the solenoid is charged from 1A1A to 2A2A in 0.1sec⁡0.1\sec .

Explanation

Solution

The voltage induced in the solenoid is directly proportional to the number of turns of coil and change in magnetic flux by applying Faraday's second law of electromagnetic induction.

Formula used:
Change in magnetic field, ΔB=μ0nΔI\Delta B = {\mu _0}n\Delta I
Where, n=n = Number of turns per unit length; Turn density
I=I = Current in the coil
μ0={\mu _0} = Magnetic Permeability in free space

EMF or voltage induced, ε=Ndϕdt\varepsilon = N\dfrac{{d\phi }}{{dt}}
Where, N=N = Number of turns in solenoid
ϕ=B×A\phi = B \times A

Complete step by step solution:
A solenoid consists of a straight coil with many turns to produce a nearly uniform magnetic field inside. The direction of the magnetic field within the coil depends upon the direction of current.
The magnetic field induced in a long solenoid is given the formula:
B=μnIB = \mu nI
where nn = Number of turns per unit length; Turn density
II = Current in the coil
μ\mu = Magnetic Permeability
We are given that the Changing current, ΔI=2−1=1A\Delta I = 2 - 1 = 1A;
Changing time, Δt=0.1sec⁡\Delta t = 0.1\sec ; Area of cross section, A=1×10−4m2A = 1 \times {10^{ - 4}}{m^2}, Number of turns, N=50×10=500N = 50 \times 10 = 500
Therefore, the required magnetic field, B=μnΔIB = \mu n\Delta I
⇒B=4×3.14×10−4×10×1\Rightarrow B = 4 \times 3.14 \times {10^{ - 4}} \times 10 \times 1
Which gives, B=1.25×10−5TB = 1.25 \times {10^{ - 5}}T
Now for calculating induced EMF we can put the value of BB in formula 2 (in formulas used) which gives ε=Ndϕdt\varepsilon = N\dfrac{{d\phi }}{{dt}}. Where, N=N = Number of turns in solenoid, ϕ=B×A\phi = B \times A

That is, ε=NB×Adt\varepsilon = N\dfrac{{B \times A}}{{dt}}
Putting the values, ε=500×1.25×10−5×1×10−40.1=500×1.25×10−10\varepsilon = 500 \times \dfrac{{1.25 \times {{10}^{ - 5}} \times 1 \times {{10}^{ - 4}}}}{{0.1}} = 500 \times 1.25 \times {10^{ - 10}}
Therefore, ε=6.25×10−6V.\varepsilon = 6.25 \times {10^{ - 6}}V.

Hence the final induced emf across the ends is 6.25μV6.25\mu V. This will be our final answer.

Note: A solenoid is like an electromagnet which is used to convert electric current into mechanical forces. The magnetic field within the solenoid depends on the density of turns and current passing through it. A solenoid is the permanent magnet that can be turned off and on with will.