Solveeit Logo
Login

Question

Question: A small loop of resistance \(2\pi \Omega \) and having cross sectional area \({10^{ - 4}}{m^2}\) is ...

A small loop of resistance 2πΩ2\pi \Omega and having cross sectional area 10−4m2{10^{ - 4}}{m^2} is placed concentrically and coplanar with a bigger loop of radius 0.1 m0.1\,m. A steady current of 1A1A is passed in a bigger loop. The smaller loop is rotated about its diameter with an angular velocity ω\omega .The value of induced current in the smaller loop will be [in ampere] :-
A. π×10−9ωsin⁡ωt\pi \times {10^{ - 9}}\omega \sin \omega t
B. π2×10−11ωsin⁡ωt\dfrac{\pi }{2} \times {10^{ - 11}}\omega \sin \omega t
C. 10−9ωsin⁡ωt{10^{ - 9}}\omega \sin \omega t
D. 10−10ωsin⁡ωt{10^{ - 10}}\omega \sin \omega t

Explanation

Solution

To calculate the value of induced current in the smaller loop, we have to use the concept of induced current in which when the magnetic flux passing through a loop changes, an induced emf and hence induced current is produced in the circuit. Also, we use the concept of magnetic field through a circular loop.

Complete step by step answer:
Let us write the data given in the question,
R=2πΩR = 2\pi \Omega , s=10−4m2s = {10^{ - 4}}{m^2}, r=0.1mr = 0.1m , I=1AI = 1A
The magnetic field through the circular loop is given by
B=μ0I2rB = \dfrac{{{\mu _0}I}}{{2r}}
Where, II - current flowing through the loop and rr -radius of the loop.
B=μ0I2r=4π×10−7×12×0.1B = \dfrac{{{\mu _0}I}}{{2r}} = \dfrac{{4\pi \times {{10}^{ - 7}} \times 1}}{{2 \times 0.1}}
⇒B=2π×10−6−−−−−−−−−(1)\Rightarrow B = 2\pi \times {10^{ - 6}} - - - - - - - - - (1)

The electromotive force or induced emf through a loop having a small area dsds is given by
ε=−dϕBdt=−d(Bscos⁡θ)dt\varepsilon = - \dfrac{{d{\phi _B}}}{{dt}} = - \dfrac{{d\left( {Bs\cos \theta } \right)}}{{dt}}
⇒ε=−2π×10−6dsˉdt−−−−−−−−−−(2)\Rightarrow \varepsilon = - 2\pi \times {10^{ - 6}}\dfrac{{d\bar s}}{{dt}} - - - - - - - - - - (2)
We have sˉ=scos⁡ωt\bar s = s\cos \omega t
dsˉdt=−sωsin⁡ωt\dfrac{{d\bar s}}{{dt}} = - s\omega \sin \omega t
Substituting in equation (2)(2) , we get
ε=−2π×10−6×(−sωsin⁡ωt)\varepsilon = - 2\pi \times {10^{ - 6}} \times \left( { - s\omega \sin \omega t} \right)
⇒ε=2πωsin⁡ωt×10−6×10−4\Rightarrow \varepsilon = 2\pi \omega \sin \omega t \times {10^{ - 6}} \times {10^{ - 4}}
⇒ε=2πωsin⁡ωt×10−10−−−−−−−(3)\Rightarrow \varepsilon = 2\pi \omega \sin \omega t \times {10^{ - 10}} - - - - - - - (3)
Now, The induced current through smaller loop is given by
i=εR ⇒i=2πωsin⁡ωt×10−102πi = \dfrac{\varepsilon }{R} \\\ \Rightarrow i= \dfrac{{2\pi \omega \sin \omega t \times {{10}^{ - 10}}}}{{2\pi }}
∴i=10−10ωsin⁡ωt\therefore i = {10^{ - 10}}\omega \sin \omega t

Hence, option D is correct.

Note: We should know the magnetic field due to a circular loop and the direction of the induced current is such that the induced emf in the loop due to changing flux always opposes the change in the magnetic flux. When the angle between θ\theta and BB changes by rotating the loop with angular frequency, the magnetic flux changes and emf is induced.