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5.Magnetism and Matter
medium
Write difference between electrostatic and magnetics :
Option A
Option B
Option C
Option D
Solution
| electrostatic | magnetics |
| $(1)$ $\frac{1}{\epsilon_{0}}\left(\epsilon_{0}=\right.$ permittivity of vacuum $)$ | $(1)$ $\mu_{0}\left(\mu_{0}=\right.$ permeability of vacuum $)$ |
| $(2)$Electric charge $q$ | $(2)$ Magnetic pole $\left(q_{m}\right)$ |
| $(3)$Electric dipole moment $\vec{p}=(2 \vec{a})(q)$ | $(3)$ Magnetic dipole moment $\vec{m}=(2 \vec{l})\left(q_{m}\right)$ |
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$(4)$Electric force between two charges $\mathrm{F}=\frac{\mu_{0}}{4 \pi} \frac{\left(q_{m 1}\right)\left(q_{m 2}\right)}{r^{2}}$ |
$(4)$ Magnetic force between two magnetic poles $\mathrm{F}=\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}$ |
|
$(5)$Electric field on the axis of electric dipole $\overrightarrow{\mathrm{E}}=\frac{2 \vec{p}}{4 \pi \epsilon_{0} r^{3}} \quad(r>>l)$ |
$(5)$ Magnetic field on the axis of bar magnet $\overrightarrow{\mathrm{B}}=\frac{\mu_{0}}{4 \pi} \frac{2 \vec{m}}{r^{3}}$ (For small magnet $r>>l$ ) |
|
$(6)$Electric field on equatorial line $\overrightarrow{\mathrm{E}}=\frac{-\vec{p}}{4 \pi \epsilon_{0} r^{3}} \quad(r>>l)$ |
$(6)$ Magnetic field on equatorial line $\overrightarrow{\mathrm{B}}=-\frac{\mu_{0}}{4 \pi} \frac{\vec{m}}{r^{3}}$ (For small magnet $r>>l$ ) |
| $(7)$Torque $\vec{\tau}=\vec{p} \times \overrightarrow{\mathrm{E}}$ | $(7)$Torque $\vec{\tau}=\vec{m} \times \overrightarrow{\mathrm{B}}$ |
| $(8)$Potential energy $\mathrm{U}=-\vec{p} \cdot \overrightarrow{\mathrm{E}}$ | $(8)$Potential energy $\mathrm{U}=-\vec{m} \cdot \overrightarrow{\mathrm{B}}$ |
| $(9)$Work done $\mathrm{W}=\mathrm{P} \varepsilon\left[\cos \theta_{1}-\cos \theta_{2}\right]$ | $(9)$Work done $\mathrm{W}=m \mathrm{~B}\left[\cos \theta_{1}-\cos \theta_{2}\right]$ |
Standard 12
Physics
