Question 1:
easy
The materials suitable for making electromagnets should have :
Question 2:
easy
The universal property among all substances is:
Question 3:
easy
Magnetic susceptibility of the following is :
Question 4:
easy
Diamagnetic substances characterise by :
Question 5:
easy
\(\varepsilon_0\) and \(\mu_0\) are the electric permittivity and magnetic permeability of free space respectively. If the corresponding quantities of a medium are \(2\varepsilon_0\) and \(1.5\mu_0\) respectively, the refractive index of the medium will nearly be
The speed of light in free space is \(c = \frac{1}{\sqrt{\varepsilon_0\mu_0}}\) and in a medium is \(v = \frac{1}{\sqrt{\varepsilon\mu}}\). The refractive index \(n = \frac{c}{v} = \sqrt{\frac{\varepsilon\mu}{\varepsilon_0\mu_0}} = \sqrt{2 \times 1.5} = \sqrt{3}\).
Question 6:
easy
A thin circular wire carrying a current \(I\) has a magnetic moment \(M\). The shape of the wire is changed to a square and it carries the same current. It will have a magnetic moment of:
For circle: \(M = I \pi R^2\) where \(2\pi R = L ⇒ R = \frac{L}{2\pi}\), so \(M = \frac{I L^2}{4\pi}\). For square of side \(a = \frac{L}{4}\): \(M' = I a^2 = \frac{I L^2}{16}\). Thus, \(M' = \frac{\pi}{4} M\).
Question 7:
easy
The magnetic materials having negative magnetic susceptibility are:
Diamagnetic substances have negative magnetic susceptibility because they develop magnetization in a direction opposite to the applied magnetic field.
Question 8:
easy
To increase the current sensitivity of a moving coil galvanometer, we should:
Current sensitivity is given by \(I_s = \frac{NBA}{C}\). To increase it, we need to increase \(N\), \(B\), \(A\) or decrease \(C\). None of the options perform these changes.
Question 9:
easy
The magnetic moment produced in a substance of \(1\text{ gm}\) is \(6 \times 10^{-7}\text{ A-m}^2\). If its density is \(5\text{ gm/cm}^3\), then the intensity of magnetisation in \(A/m\) will be:
Intensity of magnetisation is \(I = \frac{M}{V} = \frac{M\rho}{m}\). Given \(m = 1\text{ gm}\), \(M = 6 \times 10^{-7}\text{ A-m}^2\), \(\rho = 5 \times 10^3\text{ kg/m}^3\). Thus \(I = \frac{6 \times 10^{-7} \times 5 \times 10^3}{10^{-3}} = 3.0\text{ A/m}\).