A) 2B
B) 4B
C) \[B/2\]
D) B
Correct Answer: D
Solution :
| [d] Key Idea: According to Ampere's law, the line integral \[\oint{\vec{B}\,.\,d\,\vec{1}}\]. of the resultant magnetic field along a dosed, plane curve is equal to \[{{\mu }_{0}}\] times the total current crossing the area bounded by the closed curve. |
| Using Ampere's law, |
| \[\oint{\vec{B}\,.\,d\,\vec{1}}\,={{\mu }_{0}}\,({{i}_{net}})\] (i) |
| In our case, |
| \[{{i}_{net}}=\] (number of turns inside the area) |
| × (current through each turn) |
| \[=(nl)i\] |
| (n = number of turns per unit length) |
| Then, Eq, (i) can be written as, |
| \[B\ell =({{\mu }_{0}})\,(nli)\] |
| or \[B={{\mu }_{0}}\,ni\] |
| or \[B\,\propto \,\,ni\] |
| \[\therefore \] \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{{{n}_{1}}\,{{i}_{1}}}{{{n}_{2}}\,{{i}_{2}}}\] |
| Here, \[{{n}_{1}}={{n}_{1}},\,{{n}_{2}}=\frac{n}{2},\,{{i}_{1}}=i,\,{{i}_{2}}=2i,\,{{B}_{1}}=B\] |
| Hence, \[\frac{B}{{{B}_{2}}}=\frac{n}{n/2}\times \frac{i}{2i}=1\] |
| or \[{{B}_{2}}=B\] |
You need to login to perform this action.
You will be redirected in
3 sec
