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Yogita Ingle 15 hours ago
The self-inductance of the coil depends on its geometry and on the permeability of the medium. The self-induced emf is also called the back emf as it opposes any change in the current in a circuit. Physically, the self-inductance plays the role of inertia. It is the electromagnetic analogue of mass in mechanics. So, work needs to be done against the back emf (ε) in establishing the current. This work done is stored as magnetic potential energy
Posted by Ashwini Anbalagan 1 day, 2 hours ago
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Gaurav Seth 1 day, 1 hour ago
On July 7, HRD Minister Ramesh Pokhriyal announced a major CBSE syllabus reduction with 30% of the syllabus slashed for the year 2020-21 for classes 9 to 12 because of the reduction in classroom teaching time due to the Covid-19 pandemic and lockdown.
CBSE has rationalized the syllabus with the help of suggestions from NCERT and the same has been notified by a new CBSE notification as well.
Click on the given link:
<a data-toggle="collapse" href="http://cbseacademic.nic.in/Revisedcurriculum_2021.html#collapse1">Revised Senior Secondary Curriculum (XI-XII)</a>
Posted by Whether 😉 1 day, 23 hours ago
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Rounik Roy 1 day ago
Tanya 💖💖 1 day, 11 hours ago
Posted by Tanya 💖💖 2 days, 11 hours ago
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Yogita Ingle 2 days, 11 hours ago
Spherical Wavefront
- When the source of light is a point source the wavefront formed will be spherical wavefront.
- Point source means the source of light is so small that it is considered as point. It can be considered as dimensionless.
- For example: - Ripples in water are in the form of concentric circles which are spherical wavefronts.
Posted by Monica Suresh 2 days, 23 hours ago
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Gaurav Seth 2 days, 21 hours ago
Explanation:
Charge of each nucleus = 29e
∴ Net charge 2 gm sphere = (29e) × (2 × 1022) = 5.8 × 1023e c.
∴ No of electrons on sphere = 5.8 × 1023
∴ No of electrons removed to give 2μc charge
= (q/e)
= [(2 × 10–6)/(1.6 × 10–19)]
= 1.25 × 1013
∴ fraction of electrons removed = [(1.25 × 1013)/(5.8 × 1023)]
= 2.16 × 10–11
Gaurav Seth 2 days, 21 hours ago
A copper sphere of mass 2 g contains nearly 2 x 1022 atoms. The charge on the nucleus of each atom is 29 e. What fraction of the electrons must be removed from the sphere to give it a charge of +2 μC?
<hr />Total number of electrons in the sphere = 29 x 2 x 1022
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Yogita Ingle 4 days, 15 hours ago
The EM waves are produced by the accelerated charge. The electron jumping from its outer to inner orbits radiates EM waves. These EM waves are propagated as electric and magnetic fields oscillating in mutually perpendicular directions which is cause of momentum and energy.
</article>Posted by Sneha Saini 5 days, 1 hour ago
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Harsh Tiwari 12 hours ago
Gaurav Seth 4 days, 23 hours ago
Electrostatic Potential : The electrostatic potential at any point in an electric field is equal to the amount of work done per unit positive test charge or in bringing the unit positive test charge from infinite to that point, against the electrostatic force without acceleration.
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Ruhi Khan 4 days, 15 hours ago
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Yogita Ingle 5 days, 6 hours ago
In an experimental set up for photoelectric effect, the value of negative potential of collector plate or anode at which the photoelectric current will reduce to zero is called stopping potential (V0) for the given frequency of incident radiation.
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Devil 😎 5 days, 23 hours ago
Posted by Rajesh Agarwal 6 days, 13 hours ago
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Gaurav Seth 6 days, 13 hours ago
Consider a ring of radius 'a' which carries uniformly distributed positive total charge Q.
To find: electric field due to a ring at a point P lying at a distance x from its centre along the central axis perpendicular to the plane of the ring.
As the charge is distributed uniformly over the ring, the charge density over the ring is,
The perpendicular component of electric field due to charge on the ring along the x-axis cancels each other out.
As there is same charge on both sides of the ring, the magnitude of the electric field at P due to the segment of charge dQ is given by,
dE = ke 
Ex = 
1. At the centre (X = 0) , electric field is zero.
2. When x>> a, a can be neglected in the denominator.
Therefore,
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Posted by Amol Lende 1 week ago
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Gaurav Seth 1 week ago
Given an electric dipole placed in a non-uniform electric field. An electric dipole always experiences a torque when placed in uniform as well as non-uniform electric field. But in non-uniform electric field, dipole will also experience net force of attraction. So the electric dipole in non-uniform electric field experiences both torque and force.
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