A 30-turn circular coil of radius 4.00 cm and resistance 1.00 V is pla

1.	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» 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