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, then the new value of the magnetic field is

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, then the new value of the magnetic field is (A) B (B) 2B (C) 4B (D) B / 2

Tardigrade Admin · Accepted Answer

Magnetic field on the axis of a solenoid B=μ 0ni Bâ²=(μ )0((n/2))(2 i)=B

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, then the new value of the magnetic field is

$B$

$2 B$

$4 B$

$B /2$

Magnetic field on the axis of a solenoid
$B = μ_{0} ni$
$B' = (μ)_{0} (\frac{n}{2}) (2 i) = B$

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, then the new value of the magnetic field is

B

2B

4B

B/2

Magnetic field on the axis of a solenoid B=μ0​ni B′=(μ)0​(2n​)(2i)=B

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, then the new value of the magnetic field is

$B$

$2 B$

$4 B$

$B /2$

Magnetic field on the axis of a solenoid
$B = μ_{0} ni$
$B' = (μ)_{0} (\frac{n}{2}) (2 i) = B$