Question

A 30-turn circular coil of radius 4.00 cm and resistance 1.00 V is placed in a magnetic field directed perpendicular to theplane of the coil. The magnitude of the magnetic field varies in time according to the expression B 5 0.010 0t 1 0.040 0t 2,where B is in Tesla and t is in seconds. Calculate the induced emf in the coil at t 5 5.00 s

Answer 1

Given: Number of turns of the coil = 30 turns,

The radius of the coil = 4.00cm = 0.04m

Resistance = 1 ohm,

The angle between the area and vector and the magnetic field = 0°  

The given expression of the magnetic field is $B = 0.010t+0.040t^{2}$

Find: Here we have to calculate the emf in the coil at t = 5seconds

Solution:

Now, According to Faraday's Law,  

The induced emf in a closed loop is equal to the negative time of the rate of change of the magnetic flux through the loop.

$\varepsilon = - \dfrac{d\phi_{B} }{dt}$

For number of turns,

$\varepsilon = - N\dfrac{d\phi_{B} }{dt}$

The magnetic flux across the plane of area A, and in a uniform magnetic field B

$\phi_{B} = ABcos\theta$

Now, we put the value of the magnetic flux in faraday's law expression

$\varepsilon = N\dfrac{d\(BAcos\theta }{dt}$

$\varepsilon = NAcos\theta\dfrac{d\-{B} }{dt}$

Now, we put the value of the magnetic field,

$\varepsilon = NAcos\theta\dfrac{d\-{(0.010t+0.040t^{2} } )}{dt}$

Now, differentiate

$\varepsilon = NAcos\theta(0.010+0.080t)$

Now, put t = 5 seconds

$\varepsilon = NAcos\theta(0.010+0.080\times5)$

$\varepsilon = 30\times\pi\times0.04^{2} cos0^{\circ} (0.010+0.080\times5)$

$\varepsilon = 30\times3.14\times0.0016 \times1 (0.010+0.4)\\\varepsilon = 0.15072 (0.41)\\\\\varepsilon = 0.0617952V$

So, the induced emf in the coil at 5 seconds is 0.0617952V

Answer 2

Answer: 83.6 mV

Explanation: induced