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 magnetic field will be equal to

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 magnetic field will be equal to (A) B (B) 2 B (C) 4 B (D) (B/2)

Tardigrade Admin · Accepted Answer

B 1=μ0 nI B 2=μ0(( n /2))(2 I ) ⇒ B 1= B 2

Q. 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 $c m$ is halved, the new value of magnetic field will be equal to

$B$

$2 B$

$4 B$

$\frac{B}{2}$

$B_{1} = μ_{0} n I$
$B_{2} = μ_{0} (\frac{n}{2}) (2 I)$
$\Rightarrow B_{1} = B_{2}$

Q. 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 magnetic field will be equal to

B

2B

4B

2B​

B1​=μ0​nI B2​=μ0​(2n​)(2I) ⇒B1​=B2​

Q. 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 $c m$ is halved, the new value of magnetic field will be equal to

$B$

$2 B$

$4 B$

$\frac{B}{2}$

$B_{1} = μ_{0} n I$
$B_{2} = μ_{0} (\frac{n}{2}) (2 I)$
$\Rightarrow B_{1} = B_{2}$