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The magnetic flux linked with a coil (in Wb) is given by the equation
$$\phi=5t^2+3t+16$$
The magnitude of induced emf in the coil at the fourth second will be
(1) 10 V
(2) 33 V
(3) 43 V
(4) 108 V

in Physics | 36 views

Given: Magnetic flux (ϕ) = 5t2 + 3t + 16

induced emf = $$-\frac{d\phi}{dt}$$

$$\epsilon$$=$$\frac{d(5t^2+3t+16}{dt}$$ =10t+3

Therefore induced e.m.f. , when t=3,

$$|\epsilon_3|=(10\times3)+3=33 V$$

induced e.m.f. , when t=4 ,

$$|\epsilon_4|=(10\times4)+3=43 V$$

Therefor e.m.f. induced in the fourth second

=$$|\epsilon_4|-|\epsilon_3|=43-33=10V$$

by Expert (10.7k points)