B=2rμ0I.
When n turns are made, radius becomes r′.
n×2πr′=2πr⇒r′=nr.
Now, B′=2r′μ0nI=n22rμ0I=n2B.
A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be
Held on 30 Apr 2016 · Verified 9 Jul 2026.
nB
n2B
2nB
2n2B
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Two point charges +2μC and -2μC are placed 10 cm apart. The electric field at the midpoint is:
A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is :- 
The electric potential at distance r from a point charge q is V = q/(4πε₀r). The electric field E is:
To an ac power supply of 220 V at 50 Hz , a resistor of $20 \Omega$, a capacitor of reactance $25 \Omega$ and an inductor of reactance $45 \Omega$ are connected is series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively-
A wire of resistance R is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is :
Work through every NEET UG Electromagnetism PYQ, year by year.