(i) Write an expression for Biot-Savart’s law in the vector form. Write an expression for the magnetic field at the centre of a circular coil of radius R, with N turns and carrying a current I.
(ii) A helium nucleus completes one round of a circle of radius 0.8 m in 2 seconds. Find the magnetic field at the centre of the circle.
Here,
q = Charge on 2 electrons
= 2 x 1.6 x 10-19 C
= 3.2 x 10-19 C.
t = 2 seconds.
r = 0.8 m
B =
= 4π x 10-26
= 4 x 3.14 x 10-26
= 1256 x 10-25T
One of the magnets of the magnetic moment M1 is placed on one of its arms.
The readings of both the ends of aluminium pointer are noted.
The magnet is reversed end to end in the same position and the readings of both the ends of the pointer are again noted.
The magnet is now taken to the other arm and the readings are taken by keeping the magnet at the same distance d from the centre of the compass box.
The average value of these eight readings is forced out as θ1.
The experiment is repeated with the second magnet of Moment M2, keeping the magnet at the distance d from the centre of the compass box.
The average of the eight readings is noted as θ2.
Let 2l1 and 2l2 be the lengths of the two magnets. The magnetic fields produced by the magnets of moments M1 and M2 at the centre of the compass box are B1 and B2respectively.
If BH is the horizontal component of the magnetic field at the place then,
Where B and BH are two uniform magnetic fields which act at right angle to each other
on a magnet and θ is the angle which the magnet makes with the direction of BH.
Torque due to the magnetic field BH,
τ1 = mBH x MN ...(i)
Torque due to the magnetic field B,
τ2 = mB x OM ...(ii)
In equilibrium position, we have
τ1 = τ2
mB x OM = mBH x MN
B = BH x
In right angled