The permeability of free space is 4 pi Times 10^-7 T A/M. 4 pi Times 10^7 T A/M.

Question

The permeability of free space is 4 pi Times 10^-7 T A/M. 4 pi Times 10^7 T A/M. 4 pi Times 10^-7 T/A. A copper wire carries a current of 3.2 A. The number of electrons passing through the cross section of the wire each second is 2.0 Times 10^-19 2.0 Times 10^19. 2.0 Times 10^19. 3.2 Times 10^19, Two resistors of the same resistance 12 Ohm are connected in parallel. The equivalent resistance of the two resistors is 6 Ohm. 12 Ohm. 24 Ohm. A battery of internal resistance 0.2 Ohm and emf 6.0 V is put in a circuit with a charging current of 2.0 A. The terminal voltage of the battery is 6.0 V. 6.4 V. 5.6 V. The power consumed on the internal resistance is 12 W. 0.8 W. 0.4 W. A capacitor of C = 0.2 mu F initially charged to 3.0 V is discharged through a resistor. The time constant of the RC circuit is 0.6 s. The resistance of the resistor is 3.0 M Ohm. 6.0 M Ohm. 12.0 M Ohm. Two long straight parallel wires carry currents of the same magnitude 10 A in the opposite directions. They are separated by a distance d = 10 cm. The forces between them are 2.0 Times 10^-6 N/m, attractive. 2.0 Times 10^-6 N/m, repulsive, 2.0 Times 10^-6 N/m, repulsive. 2.0 Times 10^-4 N/m, attractive.

Explanation / Answer

1) Correct option is (c)

2) Correct option is (b)

i =3.2 A, q = 1.6*10^-19C, t =1 s

i = nq/t

n = it/q = 3.2*1/(1.6*10^-19) = 2*10^19

3) Correct option is (a)

R = 12*12/(12+12) = 6 ohm

4) Correct option is (b)

R =0.2 ohm , e =6 V, i =2 A

From ohms law V =iR = 2*0.2 = 0.4 V

terminal voltage = 6+0.4 =6.4 V

5) Correct option is (b)

P =i^2R = 2*2*0.2 = 0.8 W

6) correct option is (a)

C =0.2 uF, V = 3 V, T = 0.6 s

T = RC

R = T/C = 0.6s/(0.2uF) = 3 Mohms

7) Correct option is (b)

i =10 A, d = 10 cm

F/l= uoi1i2/2pi*d = (4*3.14*10^-7*10*10)/(2*3.14*0.1)

F/l = 2*10^-4 N/m, repulsive

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