
A field of strength $\dfrac{5 \times 10^4}{\pi}$ ampere turns /meter acts at right angles to a coil of 50 turns of area $10^{-2} m^2$. The coil is removed from the field in 0.1 second. Then, the e.m.f. in the coil is:
A. 0.1 V
B. 0.2 V
C. 1.96 V
D. 0.98 V
Answer
585.3k+ views
Hint: Follow the unit of magnetic field and observe that it is the value of H that has been given to us (not B). To find the e.m.f. induced we need to determine the change in flux by knowing the flux before withdrawing and after withdrawing.
Formula used:
The magnetic flux for a coil perpendicular to the magnetic field is:
$\phi = BNA$
By Faraday's law we can write the induced e.m.f. as the rate of change of flux or;
$e = - \dfrac{d \phi}{dt} = -\dfrac{\phi_2 - \phi_1}{t} $ .
The relation between B and H:
$B = \mu_0 H$
Complete step by step answer:
We are given,
Area of the coil = $10^{-2} m^2$,
Number of turns, N= 50,
Strength of the magnetic field H = $\dfrac{5 \times 10^4}{\pi}$ A/m.
We are given the unit in Ampere turns per meter which can be also denoted by AT/m. The conversion is simply:
1 AT/m = 1 A/m.
The flux for an the loop inside the magnetic field can be written as:
$\phi_1 = BNA = \mu_0 H NA$.
Keeping the given values in this we get:
$\phi_1 = \dfrac{(4 \pi \times 10^{-7}) ( 5 \times 10^4) (50) (10^{-2})}{\pi} = 10^{-2} T m^2$
As the coil is pulled out of the magnetic field, the flux becomes:
$\phi_2 = 0$
As the induced e.m.f. is written as the rate of change of flux, we may write;
$e = - \dfrac{d \phi}{dt} = \dfrac{ \phi_1}{t} $
Keeping the values in this, we get:
$e = \dfrac{ 0.01 }{0.1} = 0.1$V.
Therefore, the correct answer is option (A).
Additional Information:
The S.I. unit for magnetic induction, B is Tesla or N $A^{-1}m^{-1}$. This can be derived from the simple law of Lorentz force on current carrying conductor in a magnetic field B,
$F = BIL$.
C.G.S. unit for B is gauss or oersted and
1 gauss =$ 10^{-4} N A^{-1}m^{-1}$.
Note:
When a substance is placed inside a magnetic field, the magnetic field inside the magnetic substance is denoted by B, which tells about the magnetic flux density or magnetic induction of the substance. The magnetising field outside the substance produced by free currents is the magnetization field which is denoted by H.
Formula used:
The magnetic flux for a coil perpendicular to the magnetic field is:
$\phi = BNA$
By Faraday's law we can write the induced e.m.f. as the rate of change of flux or;
$e = - \dfrac{d \phi}{dt} = -\dfrac{\phi_2 - \phi_1}{t} $ .
The relation between B and H:
$B = \mu_0 H$
Complete step by step answer:
We are given,
Area of the coil = $10^{-2} m^2$,
Number of turns, N= 50,
Strength of the magnetic field H = $\dfrac{5 \times 10^4}{\pi}$ A/m.
We are given the unit in Ampere turns per meter which can be also denoted by AT/m. The conversion is simply:
1 AT/m = 1 A/m.
The flux for an the loop inside the magnetic field can be written as:
$\phi_1 = BNA = \mu_0 H NA$.
Keeping the given values in this we get:
$\phi_1 = \dfrac{(4 \pi \times 10^{-7}) ( 5 \times 10^4) (50) (10^{-2})}{\pi} = 10^{-2} T m^2$
As the coil is pulled out of the magnetic field, the flux becomes:
$\phi_2 = 0$
As the induced e.m.f. is written as the rate of change of flux, we may write;
$e = - \dfrac{d \phi}{dt} = \dfrac{ \phi_1}{t} $
Keeping the values in this, we get:
$e = \dfrac{ 0.01 }{0.1} = 0.1$V.
Therefore, the correct answer is option (A).
Additional Information:
The S.I. unit for magnetic induction, B is Tesla or N $A^{-1}m^{-1}$. This can be derived from the simple law of Lorentz force on current carrying conductor in a magnetic field B,
$F = BIL$.
C.G.S. unit for B is gauss or oersted and
1 gauss =$ 10^{-4} N A^{-1}m^{-1}$.
Note:
When a substance is placed inside a magnetic field, the magnetic field inside the magnetic substance is denoted by B, which tells about the magnetic flux density or magnetic induction of the substance. The magnetising field outside the substance produced by free currents is the magnetization field which is denoted by H.
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