A 5-turn square loop (10 cm along a side resistance = 4.0 Ohm) is placed in a ma

Question

A 5-turn square loop (10 cm along a side resistance = 4.0 Ohm) is placed in a magnetic field that makes an angle of 30 degree with the plane of the loop. The magnitude of this field varies with time according to B = 0.50t^2. where t is measured in s and B in T. What is the induced current in the coil at t = 4.0 s? a) 25 mA b) 5.0 mA c) 13 mA d) 43 mA e) 50 mA A 50-cm wire placed in an east-west direction is moved horizontally to the north with a speed of 2.0 m/s. The horizontal component of the earth's magnetic field at that location is 25 mu T toward the north and the vertical component is 50mu T downward. What is the emf induced between the ends of the wire? a) 10 mu V b) 20 mu V c) 30 mu V d) 40 mu V e) 50 mu V A. what me would the current in a 100-mH inductor have to change to induce an emf of 1000 V in the inductor? a) 100 A/s b) 1 A/s c) 1000 A/s d) 10,000 A/s c) 10 A/s

Explanation / Answer

Q5.
flux through the coil=number of turns*B*area*cos(angle between B and the normal vector to the plane)

=5*0.5*t^2*0.1^2*cos(90-30)

=0.0125*t^2

then induced emf magnitude=rate of change of flux

=0.025*t

induced current=induced emf/resistance

=0.025*t/4

=6.25*10^(-3)*t A

at t=4 seconds,

induced current=0.025 A

=25 mA

hence option A is correct.

Q6.

emf indcued due to magnetic field which is perpendicular.

in this case, it is the vertical component.

emf magnitude=magnetic field*length*speed

=50 uT*0.5 m*2 m/s

=50 uV

so option E is correct.

Q7.

emf in an inductor=inductance*rate of change of current

==>1000=0.1*rate of change

==>rate of change of current=10000 A/s

hence option D is correct.

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