Two circular coils X and Y have equal number of turns and carry equal currents in the same sense and subtend same solid angle at point O. If the smaller coil X is midway between O and Y, then if we represent the magnetic induction due to bigger coil Y at O as BY and that due to smaller coil X at O as Bx :
A thin flexible wire of length L is connected to two adjacent fixed points carries a current I in the clockwise direction, as shown in the figure. When system is put in a uniform magnetic field of strength B going into the plane of paper, the wire takes the shape of a circle. The tension in the wire is :
A helium nucleus is moving in a circular path of radius \(0.8\text{ m}\). If it takes \(2\text{ sec}\) to complete one revolution, the magnetic field produced at the centre of the circle is:
Current is \(I = \frac{q}{T} = \frac{2e}{2} = e = 1.6 \times 10^{-19}\text{ A}\). The magnetic field at the centre is \(B = \frac{\mu_0 I}{2R} = \frac{\mu_0 (1.6 \times 10^{-19})}{2(0.8)} = \mu_0 \times 10^{-19}\text{ T}\).