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Find the flux density due to a conductor of length 6m and carrying a current of 3A(in 10-7 order)
a Explanation: The flux density is B = μH, where H = I/2πR. Put I = 3 and R = 6, we get B= 4π x 10-7 x 3/2π x 6 = 1 x 10-7 units.
a
See lessExplanation: The flux density is B = μH, where H = I/2πR. Put I = 3 and R = 6, we get B= 4π x 10-7 x 3/2π x 6 = 1 x 10-7 units.
When currents are moving in the same direction in two conductors, then the force will be
a Explanation: When two conductors are having currents moving in the same direction then the forces of the two conductors will be moving towards each other or attractive.
a
See lessExplanation: When two conductors are having currents moving in the same direction
then the forces of the two conductors will be moving towards each other or attractive.
The force per unit length of two conductors carrying equal currents of 5A separated by a distance of 20cm in air(in 10-6 order)
a Explanation: The force per unit length of two conductors is given by F = μ I1xI2/2πD, where I1 = I2 = 5 and D = 0.2. Thus F = 4π x 10-7 x 52/ 2π x 0.2 = 25 x 10-6 units.
a
See lessExplanation: The force per unit length of two conductors is given by
F = μ I1xI2/2πD, where I1 = I2 = 5 and D = 0.2. Thus F = 4π x 10-7 x 52/ 2π x 0.2 = 25 x 10-6 units.
Find the electric field when the velocity of the field is 12m/s and the flux density is 8.75 units.
b Explanation: The electric field intensity is the product of the velocity and the magnetic flux density ie, E = v x B = 12 x 8.75 = 105 units.
b
See lessExplanation: The electric field intensity is the product of the velocity and the magnetic
flux density ie, E = v x B = 12 x 8.75 = 105 units.
Find the magnetic force when a charge 3.5C with flux density of 4 units is having a velocity of 2m/s
b Explanation: The magnetic force is given by F = q(v x B), where q = 3.5C, v = 2m/s and B = 4 units. Thus we get F = 3.5(2 x 4) = 28 units.
b
See lessExplanation: The magnetic force is given by F = q(v x B), where q = 3.5C, v = 2m/s and B = 4 units. Thus we get F = 3.5(2 x 4) = 28 units.
Find the electric force when the charge of 2C is subjected to an electric field of 6 units.
c Explanation: The electric force is given by F = qE, where q = 2C and E = 6 units. Thus we get F = 2 x 6 = 12 units.
c
See lessExplanation: The electric force is given by F = qE, where q = 2C and E = 6 units. Thus
we get F = 2 x 6 = 12 units.
Compute the power consumed by a material with current density 15 units in an area of 100 units. The potential measured across the material is 20V.
c Explanation: Power is given by, P= V X I, where I = J X A is the current. Thus power P = V X J X A = 20 X 15 X 100 = 30,000 joule = 30kJ.
c
Explanation: Power is given by, P= V X I, where I = J X A is the current. Thus power P = V X J X A = 20 X 15 X 100 = 30,000 joule = 30kJ.
See lessCalculate the power of a material with electric field 100 units at a distance of 10cm with a current of 2A flowing through it.
b Explanation: Power is defined as the product of voltage and current. P = V X I, where V = E X d. Thus P = E X d X I = 100 X 0.1 X 2 = 20 units.
b
See lessExplanation: Power is defined as the product of voltage and current.
P = V X I, where V = E X d. Thus P = E X d X I = 100 X 0.1 X 2 = 20 units.
From the formula F = qE, can prove that work done is a product of force and displacement. State True/False
a Explanation: We know that F = qE = qV/d and W = qV. Thus it is clear that qV = W and qV = Fd. On equating both, we get W = Fd, which is the required proof.
a
See lessExplanation: We know that F = qE = qV/d and W = qV. Thus it is clear that qV = W and qV = Fd. On equating both, we get W = Fd, which is the required proof.
Find the force on a conductor of length 12m and magnetic flux density 20 units when a current of 0.5A is flowing through it.
b Explanation: The force on a conductor is given by F = BIL, where B = 20, I = 0.5 and L =12. Force F = 20 X 0.5 x 12 = 120 N.
b
See lessExplanation: The force on a conductor is given by F = BIL, where B = 20, I = 0.5 and L =12. Force F = 20 X 0.5 x 12 = 120 N.