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Find the magnetic flux density when a point from a finite current length element of current 0.5A and radius 100nm.
c Explanation: The magnetic flux density is B = μH, where H is given by I/2πr. Put μ = 4π x 10-7, I = 0.5 and r = 10-7, we get B = 4π x 10-7 x 0.5/2π x 10-7 = 1 unit.
c
See lessExplanation: The magnetic flux density is B = μH, where H is given by I/2πr. Put μ = 4π
x 10-7, I = 0.5 and r = 10-7, we get B = 4π x 10-7 x 0.5/2π x 10-7 = 1 unit.
Find the magnetic field intensity at the centre O of a square of the sides equal to 5m and carrying 10A of current.
d Explanation: The magnetic field is given by H = 4I/√2πω. Put I = 10 and ω = 5m. Thus H = 4 x 10/√2π(5) = 1.8 unit.
d
See lessExplanation: The magnetic field is given by H = 4I/√2πω. Put I = 10 and ω = 5m. Thus H
= 4 x 10/√2π(5) = 1.8 unit.
The current element of the solenoid of turns 100, length 2m and current 0.5A is given by,
c Explanation: The current element of the solenoid is given by NI dx/L. Put N = 100, I = 0.5 and L = 2 to get, I dx = 100 x 0.5 x dx/2 = 25 dx.
c
See lessExplanation: The current element of the solenoid is given by NI dx/L. Put N = 100, I = 0.5
and L = 2 to get, I dx = 100 x 0.5 x dx/2 = 25 dx.
Calculate the magnetic field at a point on the centre of the circular conductor of radius 2m with current 8A.
b Explanation: The magnetic field due to a point in the centre of the circular conductor is given by H = I/2a. Put I = 8A and a = 2m, we get H = 8/4 = 2 units.
b
See lessExplanation: The magnetic field due to a point in the centre of the circular conductor is
given by H = I/2a. Put I = 8A and a = 2m, we get H = 8/4 = 2 units.
Find the magnetic field of a finite current element with 2A current and height 1/2π is
a Explanation: The magnetic field due to a finite current element is given by H = I/2πh. Put I = 2 and h = 1/2π, we get H = 1 unit.
a
See lessExplanation: The magnetic field due to a finite current element is given by H = I/2πh. Put
I = 2 and h = 1/2π, we get H = 1 unit.
Which of the following cannot be computed using the Biot Savart law?
c Explanation: The Biot Savart law is used to calculate magnetic field intensity. Using which we can calculate flux density and permeability by the formula B = μH.
c
See lessExplanation: The Biot Savart law is used to calculate magnetic field intensity. Using
which we can calculate flux density and permeability by the formula B = μH.
Biot Savart law in magnetic field is analogous to which law in electric field?
c Explanation: Biot Savart law states that the magnetic flux density H = I.dl sinθ/4πr2, which is analogous to the electric field F = q1q2/4πεr2, which is the Coulomb’s law.
c
See lessExplanation: Biot Savart law states that the magnetic flux density H = I.dl sinθ/4πr2,
which is analogous to the electric field F = q1q2/4πεr2, which is the Coulomb’s law.
Find the total capacitances when two capacitors 2F and 5F are in series.
d Explanation: Two capacitances in series gives C = C1C2/C1 + C2 = 2 x 5/2 + 5 = 10/7 farad.
d
See lessExplanation: Two capacitances in series gives C = C1C2/C1 + C2 = 2 x 5/2 + 5 = 10/7
farad.
Find the electric field due to charge density of 1/18 and distance from a point P is 0.5 in air(in 109 order)
c Explanation: The electric field for this case is given by, E = ρl/2πεd. Put ρl = 1/18 and d = 0.5. We get E = 2 x 109 units.
c
See lessExplanation: The electric field for this case is given by, E = ρl/2πεd. Put ρl = 1/18 and d
= 0.5. We get E = 2 x 109 units.
Calculate the electric field due to a surface charge of 20 units on a plate in air(in 1012 order)
b Explanation: The electric field due to plate of charge will be E = ρs/2εo. Put ρs = 20, we get E = 20/(2 x 8.854 x 10-12) = 1.129 x 1012 units.
b
See lessExplanation: The electric field due to plate of charge will be E = ρs/2εo. Put ρs = 20, we
get E = 20/(2 x 8.854 x 10-12) = 1.129 x 1012 units.