Question

A long solenoid carrying a current produces a magnetic field b along its axis. If the current is double and number of turns per cm is halved, the new value of magnetic field is (1) 4b (2) b/2 (3) b (4) 2b

Answer 1

Magnetic Field is directly proportional to current and number of turns.

So new Magnetic field=2I * 1/2N=b

Answer 2

Answer:

New value of magnetic field is b.

Explanation:

It is given that, a long solenoid carrying a current produces a magnetic field b along its axis.

We know that the magnetic field in a solenoid is given by :

$b=\mu_0nI$.......(1)

or

$b=\mu_0\dfrac{N}{L}I$

Where

$\mu_0$ is the permeability

n is the number of turns per unit length

L is the length of solenoid

I is the current flowing in the solenoid.

According to condition, the current is doubled and number of turns per cm is halved. Then the magnetic field is b'

$b'=\mu_0\times \dfrac{n}{2}\times 2I$

So, b' = b

i.e. the magnetic field does not changes. It will remains the same i.e. b. Hence, the correct option is (3) "b".