6 questions · timed · auto-graded
Explanation:
Denial of (a):
Tangent galvanometer is an instrument used to measure electric current; it cannot be used to the measure magnetic moment of a bar magnet.
Justification of (b) and (c):
Deflection magnetometer is used to measure
$\frac{\text{M}}{\text{B}_\text{H}}$ of a permanent bar magnet.Similarly, oscillation magnetometer is used to measure M BH of a bar magnet. So, if earth's horizontal field, BH, is known, then the magnetic moment of a bar magnet, M, can be measured.
Justification of (d):
Using deflection and oscillation magnetometers, we can calculate MBHMBH and M BH, respectively. Therefore, if we multiply the result obtained from both the instruments, then BH cancels out as $\frac{\text{M}}{\text{B}_\text{H}}\times\text{MB}_\text{H}=\text{M}^2$. Thus, the value of BH is not required.
Therefore, we can use both deflection and oscillation magnetometers if the earth's horizontal field is not known.
$\text{U}(\overrightarrow{\text{r}_2})-\text{U}(\overrightarrow{\text{r}_1})=-\int\limits^{\vec{\text{r}}_2}_{\vec{\text{r}_1}} \vec{\text{B}}.\text{d}\vec{\text{l}}.$
Apply this equation to a closed curve enclosing a long atraicht wire. The RHS of the above equation is then
$-{\mu}_\text{o} \text{ i}$ by Ampere's law. We see that $\text{U}(\vec{\text{r}_2})\neq\text{U}(\vec{\text{r}_1})$ even when $\vec{\text{r}_2}=\vec{\text{r}_1}.$Can we have a magnetic acalar potential in this case?A magnet always has two poles.