Question

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 that 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?

Answer 1

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.