The net magnetic moment of this system of two circular loops is :
\(μ_1 = πr^2_1 \times I_1\)
\(μ_2 = πr^2_2 \times I_2\)
Hence,\(μ_{net} = (μ_2-μ_1)(-\hat k)\)
= \(π(r^2_2-r^2_1)I(-\hat k)\)
= \(3.142 \times (0.5^2-03^2)\times 7(-\hat k)\)
= \(- \frac{7}{2}\hat k \;Am^2\)
A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
Electromagnetic Induction is a current produced by the voltage production due to a changing magnetic field. This happens in one of the two conditions:-
The electromagnetic induction is mathematically represented as:-
e=N × d∅.dt
Where