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pashok25 [27]
4 years ago
14

An 30-turn coil has square loops measuring 0.341 m along a side and a resistance of 3.61 Ω. It is placed in a magnetic field tha

t makes an angle of 37.5° with the plane of each loop. The magnitude of this field varies with time according to B = 1.45t^3, where t is measured in seconds and B in teslas. What is the induced current in the coil at t = 1.73 s?
Physics
1 answer:
Verdich [7]4 years ago
8 0

Answer:

Thus, the induced current in the coil at t =1.73s is 9.98 A.

Explanation:

Faraday's law says

$\varepsilon = N \frac{d \Phi_B}{dt} $

where N is the number of turns and \Phi_B is the magnetic flux through the square coil:

Now,

N = 30

\theta = 37.5^o

A = (0.341m)^2= 0.11623m^2

B = 1.45t^3;

therefore,

$\varepsilon = N \frac{d \Phi_B}{dt}  = N\frac{d ( BA\:cos(\theta))}{dt}  = 30*\frac{d ( (1.45t^2)(0.1163)\:cos(37.5^o))}{dt}$

=30*(0.1163)\:cos(37.5^o)*1.45*\dfrac{d ( t^3)}{dt}  = 12.04t^2

\boxed{\varepsilon = 12.04t^2}

is the emf induced in the coil.

Now, the loop is connected to R = 3.61\Omega resistance; therefore, at t = 1.73s

\varepsilon = RI = 12.04t^2

RI = 12.04(1.73)^2

RI = 36.03

I = \dfrac{36.03}{3.61\Omega }

\boxed{I = 9.98A }

Thus, the current in the coil at t =1.73s is 9.98 A.

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