Question

The magnetic field inside a solenoid of radius R is B. If the current through it is doubled and its radius is also doubled keeping the number of turns per unit length the same, the magnetic field produced by it will be:
A. $\dfrac{B}{4}$
B. $\dfrac{B}{2}$
C. 2B
D. 4B

Answer 1

Hint: As a first step, you could recall the expression for magnetic field inside a solenoid. Thus you will understand the dependency of magnetic field strength on the current flowing through it and also its radius. Now find the magnetic field for the second part by applying the necessary conditions given and thus find the answer.

Formula used: Magnetic field,
$B={{\mu }_{0}}ni$

Complete step by step answer:
In the question, we are given the magnetic field inside a solenoid of radius R as B. Now the current passing through it and the radius of the solenoid are doubled and the number of turns per unit length is the kept constant. We are supposed to find the magnetic field B’ produced by the solenoid now.
In order to answer the given question, let us recall the expression for magnetic field produced inside a solenoid which is given by,
$B={{\mu }_{0}}ni$
Where, n is the number of turns in the solenoid and $i$ is the current passing through the solenoid.
Now in the second case, the current passing through the solenoid and its radius are both doubled by keeping the number turns in it constant, so, the magnetic field will be given by,
$B'={{\mu }_{0}}n\left( 2i \right)$
$\therefore B'=2B$
Therefore, we found that the magnetic field doubled when the current doubled as the magnetic field directly proportional to the current passing through it.

So, the correct answer is “Option C”.

Note: From the recalled expression for magnetic field inside a solenoid, we see that the magnetic field is independent of the radius of the solenoid. That is why doubling the radius didn’t have any effect on the value magnetic field found in the second part. Also, the field strength is also independent of the position inside the solenoid.