To compute the power consumed by a material, we can use the formula for power ((P)):[P = I times V]where:- (I) is the current in amperes (A)- (V) is the potential difference in volts (V)Given:- Current density ((J)) is 15 units (assuming the units are (text{A/m}^2))- Area ((A)) is 100 units ((text{mRead more
To compute the power consumed by a material, we can use the formula for power ((P)):
[P = I times V]
where:
– (I) is the current in amperes (A)
– (V) is the potential difference in volts (V)
Given:
– Current density ((J)) is 15 units (assuming the units are (text{A/m}^2))
– Area ((A)) is 100 units ((text{m}^2), assuming based on the units given for current density)
– Potential ((V)) across the material is 20V
First, we need to find the total current ((I)) flowing through the material. The current ((I)) can be found using the formula:
[I = J times A]
Substituting the given values:
[I = 15 times 100 = 1500, text{A}]
Now, using the power formula:
[P = I times V]
[P = 1500 times 20 = 30000, text{W}]
Therefore, the power consumed by the material is 30,000 Watts (or 30 kW).
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Find the potential due the dipole when the angle subtended by the two charges at the point P is perpendicular.
The electric potential (V) due to a dipole at a point in space is given by the equation:[ V = frac{k cdot p cdot cos(theta)}{r^2} ]where:- (k) is the Coulomb's constant ((8.987 times 10^9 , text{Nm}^2/text{C}^2)),- (p) is the magnitude of the dipole moment ((q cdot d), where (q) is the charge magnitRead more
The electric potential (V) due to a dipole at a point in space is given by the equation:
[ V = frac{k cdot p cdot cos(theta)}{r^2} ]
where:
– (k) is the Coulomb’s constant ((8.987 times 10^9 , text{Nm}^2/text{C}^2)),
– (p) is the magnitude of the dipole moment ((q cdot d), where (q) is the charge magnitude and (d) is the separation distance between the charges),
– (theta) is the angle between the dipole moment vector and the line joining the point in space to the midpoint of the dipole,
– (r) is the distance from the midpoint of the dipole to the point in space where the electric potential is being calculated.
When the angle ((theta)) subtended by the two charges at the point (P) is perpendicular, i.e., (theta = 90^circ), (cos(90^circ) = 0). Therefore, the potential (V) at point (P) due to the dipole, when (theta = 90^circ), is:
[ V = frac{k cdot p cdot cos(90^circ)}{r^2} = frac{k cdot p cdot 0}{r^
See lessFind the angle at which the potential due a dipole is measured, when the distance from one charge is 12cm and that due to other is 11cm, separated to each other by a distance of 2cm.
To find the angle at which the potential due to a dipole is measured, given the distances from the charges, consider a dipole consisting of charges (+q) and (-q), separated by a distance (2a). The position where the potential is being measured forms a triangle with the line joining the charges. GiveRead more
To find the angle at which the potential due to a dipole is measured, given the distances from the charges, consider a dipole consisting of charges (+q) and (-q), separated by a distance (2a). The position where the potential is being measured forms a triangle with the line joining the charges. Given that the distances from the positive and negative charges are 12 cm and 11 cm, respectively, and the separation between the charges ((2a)) is 2 cm, we can find the angle using the law of cosines.
Let us denote:
– (r_+) as the distance from the positive charge ((12 cm)),
– (r_-) as the distance from the negative charge ((11 cm)),
– (d = 2 cm) as the distance between the charges,
– (theta) as the angle the position vector (from the negative to the positive charge) makes with the perpendicular bisector of the dipole.
Using the law of cosines on the triangle formed by the position of measurement and the two charges, we have:
[r_+^2 = r_-^2 + d^2 – 2 cdot r_- cdot d cdot cos(theta)]
Substituting values:
[12^2 = 11^2 + 2^2 – 2 cdot 11 cdot 2 cdot cos(theta)]
S
See lessCalculate the dipole moment of a dipole with equal charges 2C and -2C separated by a distance of 2cm.
The dipole moment ((p)) is calculated using the formula:[p = q times d]where (q) is the magnitude of the charge and (d) is the separation distance between the charges.Given:- (q = 2C) (Note: In a real-world context, a charge of 2 Coulombs is extremely large for discussions of dipole moments, suggestRead more
The dipole moment ((p)) is calculated using the formula:
[p = q times d]
where (q) is the magnitude of the charge and (d) is the separation distance between the charges.
Given:
– (q = 2C) (Note: In a real-world context, a charge of 2 Coulombs is extremely large for discussions of dipole moments, suggesting this might be a hypothetical question for illustrative purposes.)
– (d = 2cm = 0.02m) (It’s important to convert distance to meters in the SI system for consistency.)
Substituting the given values into the formula:
[p = 2C times 0.02m = 0.04 Ccdot m]
So, the dipole moment of the dipole is (0.04) Coulomb-meter ((Ccdot m)).
See lessCompute 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
To compute the power consumed by the material, we can use the formula for power ((P)) in electrical systems, which is given by (P = IV), where (I) is the current in amperes (A) and (V) is the potential difference across the material in volts (V).However, to find the current ((I)) flowing through theRead more
To compute the power consumed by the material, we can use the formula for power ((P)) in electrical systems, which is given by (P = IV), where (I) is the current in amperes (A) and (V) is the potential difference across the material in volts (V).
However, to find the current ((I)) flowing through the material, we first need to determine the total current from the given current density ((J)). Current density is defined as the current flow per unit area, given by (J = I/A), where (J) is the current density, (I) is the total current, and (A) is the area through which the current flows.
Given:
– Current density ((J)) = 15 units (assuming units are (A/m^2) for the purpose of this calculation, as actual units were not specified),
– Area ((A)) = 100 units ((m^2), assuming meters squared for coherence with current density units),
– Potential difference ((V)) = 20V.
We find the total current flowing ((I)) first:
[J = frac{I}{A} implies I = J times A]
[I = 15 , A/m^2 times 100 , m^2 = 1500 , A]
Then, we substitute (I) into
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.
To calculate the power of a material given an electric field (E), distance (d), and current (I) flowing through it, we use the formula relating electric field strength, distance, and potential difference (V), and then apply the power formula.First, the electric field (E) is given in units (assumed tRead more
To calculate the power of a material given an electric field (E), distance (d), and current (I) flowing through it, we use the formula relating electric field strength, distance, and potential difference (V), and then apply the power formula.
First, the electric field (E) is given in units (assumed to be Volts per meter, V/m), the distance (d) is given in centimeters (10 cm, which needs to be converted to meters, 10 cm = 0.1 m for the formula to work correctly since standard SI units are necessary), and the current (I) is given in Amperes (A).
The electric field is defined as the electric force per unit charge, and its relationship with potential difference (voltage, V) across a distance (d) is given by:
[ V = E times d ]
Substituting the given values:
[ V = 100 , text{V/m} times 0.1 , text{m} = 10 , text{V} ]
Now, with the voltage calculated, we can find the power (P) using the formula:
[ P = V times I ]
where ( P ) is the power in watts (W), ( V ) is the potential difference in volts (V), and ( I ) is the current in amperes (A).
[ P = 10 ,
See lessFind the force on a conductor of length 12m and magnetic flux density 20 units when a current of 0.5A is flowing through it.
To find the force on a conductor within a magnetic field, we use the formula:[ F = B I L sin(theta) ]Where:- (F) is the force (in newtons, N),- (B) is the magnetic flux density (in teslas, T; in this case, you mentioned units, so I'll assume the units are meant to be teslas),- (I) is the current (inRead more
To find the force on a conductor within a magnetic field, we use the formula:
[ F = B I L sin(theta) ]
Where:
– (F) is the force (in newtons, N),
– (B) is the magnetic flux density (in teslas, T; in this case, you mentioned units, so I’ll assume the units are meant to be teslas),
– (I) is the current (in amperes, A),
– (L) is the length of the conductor (in meters, m),
– (theta) is the angle between the direction of the current and the direction of the magnetic field.
Given:
– (B = 20) tesla (assuming the “units” mentioned are tesla),
– (I = 0.5) A,
– (L = 12) m,
– Assuming the angle (theta = 90^circ) (since the angle isn’t provided, and the maximum force occurs when the angle is 90 degrees, which means (sin(90^circ) = 1)).
Plugging in the values:
[ F = 20 times 0.5 times 12 times sin(90^circ) ]
[ F = 10 times 12 ]
[ F = 120 text{ N} ]
Thus, the force on the conductor is 120 newtons.
See lessFind the current in a conductor with resistance 2 ohm, electric field 2 units and distance 100cm
To find the current in the conductor, we need to understand the relationship between the given quantities. The electric field (E) in a conductor where an electric potential (V) is applied across a distance (d) can be expressed as (E = V / d). However, the electric field is given as 2 units (assumingRead more
To find the current in the conductor, we need to understand the relationship between the given quantities. The electric field (E) in a conductor where an electric potential (V) is applied across a distance (d) can be expressed as (E = V / d). However, the electric field is given as 2 units (assuming SI units, this would be volts per meter), and the distance is 100 cm (which is 1 meter). From this, we can directly find the potential difference (V) across the conductor since (V = E times d).
Given:
– Electric field (E) = 2 V/m
– Distance (d) = 100 cm = 1 m
First, let’s find the potential difference (V):
[ V = E times d = 2 , text{V/m} times 1 , text{m} = 2 , text{V} ]
Now, Ohm’s law states that (V = IR), where (V) is the voltage across the conductor, (I) is the current through the conductor, and (R) is the resistance of the conductor. Given the resistance (R) is 2 ohms, we rearrange Ohm’s law to solve for the current (I):
[ I = frac{V}{R} ]
Substituting the given values:
[ I = frac{2 ,
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