Advertisements
Advertisements
प्रश्न
उत्तर
Given:
`chi` = 2.1 × 10−5
To find:
% increase in the magnetic field = ?
We know the magnetic field inside the toroid without lithium
B0 = μ0nI
B0 = μ0H ...(i)
The magnetic field inside the toroid with lithium
B = µH ...(ii)
∴ B − B0 = μH − μ0H
= (μ − μ0)H
∴ The percentage increase in the magnetic field after inserting lithiu m is
% increase in magnetic field
= `(B - B_0)/B_0 xx 100`
= `((mu - mu_0)H)/(mu_0H) xx 100`
% increase in magnetic field
= `(mu - mu_0)/mu_0 xx 100`
But `(mu - mu_0)/mu_0 = chi`
∴ % increase in magnetic field
= `chi xx 100 = 2.1 xx 10^-5 xx 100`
= 0.0021%
APPEARS IN
संबंधित प्रश्न
A toroid of a central radius of 10 cm has windings of 1000 turns. For a magnetic field of 5 × 10-2 T along its central axis, what current is required to be passed through its windings?
What is Solenoid?
What is Toroid?
A solenoid of length 50 cm of the inner radius of 1 cm and is made up of 500 turns of copper wire for a current of 5 A in it. What will be the magnitude of the magnetic field inside the solenoid?
Using Ampere's Law, derive an expression for the magnetic induction inside an ideal solenoid carrying a steady current.
The length of solenoid is I whose windings are made of material of density D and resistivity p. The winding resistance is R. The inductance of solenoid is
[m = mass of winding wire, µ0 = permeability of free space]
A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 2,000 turns of a wire are wound. If the current in the wire is 10 A, the magnetic field inside the core of the toroid will be ______
Two current-carrying coils have radii r and 4r and have same magnetic induction at their centres. The ratio of voltage applied across them is ______.
Two toroids 1 and 2 have total number of turns 400 and 200 respectively with average radii 40 cm and 20 cm respectively. If they carry same current I, the ratio of the magnetic fields along the two loops is, ____________.
The magnetic induction along the axis of a toroidal solenoid is independent of ______.
A solenoid of 1.5 m length and 4 cm diameter possesses 20 turns per m. A current of 6 A is flowing through it. The magnetic induction at axis inside the solenoid is ____________.
A proton is projected with a uniform velocity 'v' along the axis of a current carrying solenoid, then ____________.
A toroid is a long coil of wire wound over a circular core. If 'r' and 'R' are the radii of the coil and toroid respectively, the coefficient of self-induction of the toroid is (The magnetic field in it is uniform and R > > r) ____________.
(N = number of turns of the coil and µ0 = permeability of free space)
A charged particle carrying a charge 'q' and moving with velocity V, enters into a solenoid carrying a current T, along its axis. If 'B' is the magnetic induction along the axis of a solenoid, then the force 'F' acting on the charged particle will be ____________.
A toroid has a core of inner radius 20 cm and outer radius 22 cm around which 4200 turns of a wire are wound. If the current in the wire is 10 A. What is the magnetic field inside the core of toroid?
In a current-carrying long solenoid, the field produced does not depend upon ______
A conducting rod along the equator is 1 m long and carries a current of 15 A from east to west. The magnitude of Earth's magnetic field at the equator is `4/3 xx 10^(-4)` T. The magnitude and direction of the force on the rod are ______.
For a solenoid and a toroid, the number of turns per unit length is n and the respective interior volume is V. The self inductance is proportional to n2 and V for ______.
A current of 10 A passes through a coil having 5 turns and produces magnetic field at the centre of the coil having magnitude 0.5 x 10-4 T. Calculate diameter of the coil.
(µ0 = 4π x 10-7 Wb/Am)
