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Question: A Coil of resistance \[10\,\Omega \] and 1000 turns have the magnetic flux line of\[5.5 \times {10^{...

A Coil of resistance 10 Ω10\,\Omega and 1000 turns have the magnetic flux line of5.5×10−4 Wb5.5 \times {10^{ - 4}}\,Wb. If the magnetic flux changed to 5×10−4 Wb5 \times {10^{ - 4}}\,Wb in 0.1 sec, then the induced charge in coil is
A. 50 C50\,C
B. 5 C5\,C
C. 2 C2\,C
D. 20 C20\,C

Explanation

Solution

We know that the current is the wire is the rate of flow of charge per unit time. The current in the coil can be found using Ohm’s law. The induced emf in the coil is due to change in magnetic flux. Use the formula for induced emf in the coil of N turns.

Formula used:
e=−NΔϕΔte = - N\dfrac{{\Delta \phi }}{{\Delta t}}, where N is the number of turns.

Complete step by step answer: We know that current in the coil is charge per unit time. Therefore,
I=qtI = \dfrac{q}{t}
⇒q=It\Rightarrow q = It

According to Ohm’s law, we can write the current in the coil as, I=eRI = \dfrac{e}{R}, where, e is the induced emf in the coil and R is the resistance of the coil. Thus, the charge in the coil is,
q=(eR)tq = \left( {\dfrac{e}{R}} \right)t …… (1)

Therefore, we need to determine the induced emf in the coil to calculate the charge.

We know that, change in flux in the coil induces emf in the coil. The induced emf in the coil is given as,
e=−NΔϕΔte = - N\dfrac{{\Delta \phi }}{{\Delta t}}
⇒e=−N(ϕ2−ϕ1Δt)\Rightarrow e = - N\left( {\dfrac{{{\phi _2} - {\phi _1}}}{{\Delta t}}} \right)

Here, N is the number of turns of the coil, ϕ2{\phi _2} is the final magnetic flux and ϕ1{\phi _1} is the initial flux.

The negative sign implies induced emf oppose the change in flux.
e=N(ϕ1−ϕ2Δt)e = N\left( {\dfrac{{{\phi _1} - {\phi _2}}}{{\Delta t}}} \right)

Substitute 1000 for N, 5.5×10−4 Wb5.5 \times {10^{ - 4}}\,Wb for ϕ1{\phi _1}, 5×10−4 Wb5 \times {10^{ - 4}}\,Wb for ϕ2{\phi _2} and 0.1 sec for Δt\Delta t in the above equation.
e=(1000)((5.5×10−4)−(5×10−4)0.1 s)e = \left( {1000} \right)\left( {\dfrac{{\left( {5.5 \times {{10}^{ - 4}}} \right) - \left( {5 \times {{10}^{ - 4}}} \right)}}{{0.1\,s}}} \right)
⇒e=(1000)(50×10−4)\Rightarrow e = \left( {1000} \right)\left( {50 \times {{10}^{ - 4}}} \right)
⇒e=5 V\Rightarrow e = 5\,V

Substitute 5 V for e, 10 Ω10\,\Omega for R and 0.1 sec for t in equation (1).
q=(5 V10 Ω)(0.1 s)q = \left( {\dfrac{{5\,V}}{{10\,\Omega }}} \right)\left( {0.1\,s} \right)
⇒q=0.05 C\Rightarrow q = 0.05\,C

So, the correct answer is option (B).

Note: Ohm’s law is applicable for the current flowing through the coil. The voltage in Ohm's law is now induced emf in the coil. While solving these types of questions, if you get the induced emf as negative, take the magnitude of it for calculation purposes. The negative sign for the induced emf implies that the induced emf oppose the magnetic flux in the coil.