Question
A long solenoid carrying a current produces a magnetic field $$B$$ along its axis. If the current is doubled and the number of turns per $$cm$$ is halved, the new value of the magnetic field is
A.
$$2B$$
B.
$$4B$$
C.
$$\frac{B}{2}$$
D.
$$B$$
Answer :
$$B$$
Solution :
For a solenoid magnetic field is given by $$B = {\mu _0}ni$$
where, $$n =$$ number of turns per unit length and
$$i =$$ current through the coil
or so for two different cases $$B \propto ni$$
$$\therefore \frac{{{B_1}}}{{{B_2}}} = \frac{{{n_1}{i_1}}}{{{n_2}{i_2}}}$$
Here, $${n_1} = n,\,{n_2} = \frac{n}{2},$$
$${i_1} = i,{i_2} = 2i,{B_1} = B$$
Hence, $$\frac{B}{{{B_2}}} = \frac{n}{{\frac{n}{2}}} \times \frac{i}{{2i}} = 1\,\,{\text{or}}\,\,{B_2} = B$$