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Question: It is given that an insulated copper wire having hundred turns has been wrapped around an iron cylin...

It is given that an insulated copper wire having hundred turns has been wrapped around an iron cylinder of area 1×10−3m21\times {{10}^{-3}}{{m}^{2}} and are connected to a resistor. The total resistance in the circuit has been given as 10Ω10\Omega . When the longitudinal magnetic induction in the iron varies from 1Wbm−21Wb{{m}^{-2}}, in the direction to 1Wbm−21Wb{{m}^{-2}}in the opposite direction, how much will the charge flow through the circuit?
A.2×10−2C B.2×10−3C C.2×10−4C D.2×10−5C \begin{aligned} & A.2\times {{10}^{-2}}C \\\ & B.2\times {{10}^{-3}}C \\\ & C.2\times {{10}^{-4}}C \\\ & D.2\times {{10}^{-5}}C \\\ \end{aligned}

Explanation

Solution

The variation in the charge flowing through the circuit will be equivalent to the ratio of the variation in the flux to the resistance connected in the circuit. The variation in the flux will be equivalent to the product of the number of turns, area of the iron cylinder, and the variation happening for the magnetic flux. This will help you in answering this question.

Complete step-by-step solution
The variation in the charge flowing through the circuit will be equivalent to the ratio of the variation in the flux to the resistance connected in the circuit. This can be written as an equation given as,
dQ=dϕRdQ=\dfrac{d\phi }{R}
The variation in the flux will be equivalent to the product of the number of turns, area of the iron cylinder, and the variation happening for the magnetic flux. This can be written as,
dϕ=nAdBd\phi =nAdB
Substituting this in the equation mentioned above can be shown as,
dQ=nAdBRdQ=\dfrac{nAdB}{R}
The charge in the circuit will be equivalent to the equation,
Q=nAΔBRQ=\dfrac{nA\Delta B}{R}
It has been already mentioned in the question that,
The number of turns will be equivalent to,
n=100n=100
The area of the iron cylinder will be,
A=1×10−3m2A=1\times {{10}^{-3}}{{m}^{2}}
The change in the magnetic field will be written as,
ΔB=2\Delta B=2
And the resistance will be equivalent to,
R=10ΩR=10\Omega
Substituting the values in the equation can be shown as,
Q=100×1×10−3×210=2×10−2CQ=\dfrac{100\times 1\times {{10}^{-3}}\times 2}{10}=2\times {{10}^{-2}}C
Therefore the charge flowing in this has been calculated. The correct answer has been mentioned as option A.

Note: A magnetic field can be defined as a vector field which is defining the magnetic influence on the motion of electric currents, electric charges, and the magnetized substances. A charge which is in motion in a magnetic field will experience a force, which will be perpendicular to its own velocity and to the magnetic field.