Question

List the factors on which the strength of magnetic field produced by
(A) A current carrying straight conductor
(B) Current carrying solenoid depends.

Answer 1

Hint For a current carrying conductor, the magnetic field is directly proportional to the current and inversely proportional to the distance from the conductor. For the solenoid, it is directly proportional to the current and number of turns per unit length of the solenoid.
Formula used: In this solution we will be using the following formula;
$\Rightarrow B={\displaystyle \frac{\mu I}{2\pi r}}$ where $B$ is the magnetic field or magnetic flux density produces by a current carrying straight conductor,$\mu$is the permeability of free space, $I$ is the current flowing through a straight conductor and $r$ is the distance from the wire considered.
$\Rightarrow B=\mu nI$ where $B$, in this case, is the magnetic field within the turns of a solenoid, $n$ is the number of turns per unit length made by the solenoid, and $I$ is the current flowing through the solenoid.

Complete step by step answer
For a), When current is flowing through a straight conductor, it produces a magnetic field which circulates about the conductor according to the right hand rule, and is given by
$\Rightarrow B={\displaystyle \frac{\mu I}{2\pi r}}$ where $\mu$ is the permeability of free space, $I$ is the current flowing through a straight conductor and $r$ is the distance from the wire considered.
Hence, from the right hand side of the expression, we see that the factors which determine the strength of the field produced by a current carrying conductor is current and the distance from the current
Thus, the answer is distance and current
For B) Magnetic field of the solenoid within its turns is given by
$\Rightarrow B=\mu nI$ where ,$n$ is the number of turns per unit length made by the solenoid.
Hence, the factors are current and number of turns per unit length.

Note
For the solenoid, we should note that the magnetic field expression is only valid within the turns of a solenoid. The magnetic field beyond the turns of a solenoid decreases rapidly to zero such that, in most cases, it is considered negligible, thus taken as zero.