A 5.0-cm-diameter coil has 20 turns and a resistanceof 0.50 ohms. A
magnetic field perpendicular to the coil is B =0.02t + 0.010t^2,
where B is in tesla and tis in seconds. Find an expression for the
induced current I(t)as a function of time.

## Answer

(I)=((N*pi*r^2)/R)*dB/dt

N=number of turns

r=half your diameter in meters

R=the resistance

dB/dt=is the derivative of the field in relation
tot=.02+.02t

=

According to Faraday's law, the relation between the induced emf and magnetic field is given as: dt -Nd(B.A) dt Here, D is the magnetic flux induced, B is the magnetic field and A is the area .d(B A dt d(В-А IR -N dt Rearrange for I as follow: ,d (В-А) 1 -N I dt R d(B) -N A dt d(B -N dt

The magnetic field acting is time differentiated: d(B) N (A) dt I R d(0.020+0.010) N (zR)| dt R d(0.02t+0.0102)T 20 (7(2.5x10 m) dt 0.5 0.0785(0.02 0.02t) -0.00157 (1+)A The current is expressed as: 0.00157 (1+t)A

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