Recent content by harsh22902

Understood it now, thank you !

Also, the term I2*R that is actually the 'net' emf , right?

The current I2 in the equation , is it due to emf induced by mutual inductance?

In the given question they have not provided the dimensions of the coil so I assumed it to be very close to the wire and having negligible dimensions compared to the wire . Then EMF induced in the coil due to the current in the wire comes out as M*a . Which when divided by resistance gives...

That clears it for me , thank you !

My approach to this problem was pretty simple and straight. There is no zero error here. 8 divisions of VS(Vernier scale)= 5 divisions on MS (Main Scale) hence 1 VS division = 5/8 mm . Now , MSR = 35 mm ( as zero of VS lies to the left of 36) Least count = 1mm - 5/8 mm = 3/8 mm Given in...

According to this definition I am unable to understand why does surface tension acts tangentially to surface of contact of liquid and capillary tube. And is the force of surface tension balancing the adhesive forces which lead to capillary rise OR it is the reason behind the capillary rise?

This is an excerpt from a high school physics textbook. I don't understand the possible reason behind this statement. If we change something in the circuit say for example add a resistor , the current and hence the flux should change. Then why/how is this statement true?

Okay so I think its clear to me now. What I was stuck on was the shielding part . Big conceptual error. Thanks! :D

I thought about it but what still appears confusing is how exactly will the induced charge due to the electric field exert a force on the charge within because for that we must have an idea about the distribution of induced charge(due to the field) on the outside and inside of the hollow sphere...

What I tried here is nothing mathematical. The charges in a cavity are sheilded from outside electric fields due to the conducting material. So what I could make out of it is that the forces acting on the charge myst be due to the induced charges on the internal cavity surface only and hence the...

Could'nt solve it

Hey folks. I am a new guy here and I like physics. Thanks.