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Question: A magnetic field of flux density \(10\;{\text{T}}\) acts normal to a coil of \(50\) turns having \(1...

A magnetic field of flux density 10  T10\;{\text{T}} acts normal to a coil of 5050 turns having 100  cm2100\;{\text{c}}{{\text{m}}^{\text{2}}} area. The e.m.f induced if the coil is removed from magnetic field in 0.10.1 second is
A) 50  V50\;{\text{V}}
B) 60  V60\;{\text{V}}
C) 80  V80\;{\text{V}}
D) 40  V40\;{\text{V}}

Explanation

Solution

Hint
The induced e.m.f is the rate of change of flux per unit time and the flux depends on the magnetic field, the area of cross section and the number of turns in the coil.

Complete step by step solution
Given the magnetic field,B=10  TB = 10\;{\text{T}}
Number of turns,N=50N = 50
Area of cross section of each coil, A=100  cm2=0.01  m2A = 100\;{\text{c}}{{\text{m}}^{\text{2}}} = 0.01\;{{\text{m}}^{\text{2}}}
And the rate of change in time, dt=0.1  sdt = 0.1\;{\text{s}}
The expression for finding the magnetic flux is given as,
ϕ=NBAcosθ\phi = NBA\cos \theta
Where, NN is the number of turns, BB is the magnetic field, AA is the area of cross section and θ\theta is the angle between magnetic field lines and the perpendicular of the area of cross section.
Let’s find the flux when the magnetic field is present.
ϕ1=NBAcosθ{\phi _1} = NBA\cos \theta
Here the magnetic field acts normal to the coil. Therefore θ=0\theta = 0 , cosθ=cos0=1\cos \theta = \cos 0 = 1 .
Substituting all the values in the expression,
ϕ1=50×10  T×0.01  m2×1 =5  weber{\phi _1} = 50 \times 10\;{\text{T}} \times 0.01\;{{\text{m}}^{\text{2}}} \times 1 \\\ = 5\;{\text{weber}}
The magnetic flux when the magnetic field present is 5  weber5\;{\text{weber}}.
When the magnetic field is removed there is no magnetic flux present.
Therefore, ϕ2=0{\phi _2} = 0 .
The induced emf is the negative of the rate of change of magnetic flux. The expression is given as,
E=dϕdt =(ϕ2ϕ1)dt  E = - \dfrac{{d\phi }}{{dt}} \\\ = - \dfrac{{\left( {{\phi _2} - {\phi _1}} \right)}}{{dt}} \\\
Substituting the values in the above expression,
E=(05)0.1  s =50  V  E = - \dfrac{{\left( {0 - 5} \right)}}{{0.1\;{\text{s}}}} \\\ = 50\;{\text{V}} \\\
The e.m.f induced if the coil is removed from the magnetic field in 0.10.1 second is 50  V50\;{\text{V}} .
Thus the answer is option (A).

Note
The emf is induced in each turn of the coil. Therefore the total induced emf is the number of turns multiplied by the emf induced in each turn. The induced emf is negative of the change of flux in unit time.