Question

A 20-cm long solenoid consists of 100 turns of a coil of radius r = 3.0 cm. A current of Io in the coiled wire produces a magnetic field of Bo in the middle of the solenoid. If we wish to produce a field that is 4-times larger, which of the following changes could work?
i) cut the radius of the coil to one-half
ii) quadruple the number of turns in the 20-cm long solenoid
iii) half the length of the 100-turn solenoid but maintain 100 turns across that length.
iv) quadruple the length of the 100-turn solenoid, but also double the radius of the coil.
v) half the length of the solenoid, and double the number of turns to 200 total turns
vi) double the current in the wire, and double the number of turns in the 20-cm long solenoid

Answer 1

Answer:

vi) Double the current in the wire, and double the number of turns in the 20-cm long solenoid

Explanation:

The magnetic field inside the solenoid and the current flowing in the coil of solenoid are related to each other by the following equation

B₀=μ₀nI₀

Where,

B₀ is the magnetic field in the middle of solenoid

n is the number of turns in the coil of solenoid

I₀ is the current flowing in the coil of solenoid

In the above equation, as μ₀ is a constant so the magnetic field will be directly proportional to the number of turns multiplied by the current. So, changing the radius of the coil or length of the coil will have no effect on the magnetic field.

As we have to increase the magnetic field by 4 times, we need to double the current as well as the number of turns as mentioned in the option vi.

Answer 2

Answer:

Quadruple the number of turns in the 20-cm long solenoid

Double the current in the wire, and double the number of turns in the 20-cm long solenoid

Quadruple the length of the 100-turn solenoid, but also double the radius of the coil.

Explanation:

The expression for the magnetic field in the middle of the solenoid is as follows;

$B_{0}=\frac{\mu NI_{0}}{L}$                                                   ........ (1)

Here, $B_{0}$ is the magnetic field in the middle of the solenoid, $\mu$ is the magnetic permeability, $I_{0}$ is the current, N is the number of turns and L is the length of the solenoid.

According to the given problem, if we wish to produce a field that is 4-times larger,

Consider the option (i) by using equation (1).

r is not present in the option. There will be no change in the magnetic field Therefore, the option (i) won't wrong.

Consider the option (ii) by using equation (1).

$B'_{0}=\frac{\mu(4N)I_{0}}{L}$

Divide this expression by the equation (1) by putting N= 4N.

$B'_{0}=4B_{0$

Therefore, the option (ii) will work.

Consider the option (iii) by using equation (1).

$B'_{0}=\frac{\mu NI_{0}}{L'}$  

Divide this expression by the equation (1) by putting $L'=\frac{L}{2}$.

$B'_{0}=2B_{0$

Therefore, the option (iii) won't work.

Consider the option (iv) by using equation (1).

$B_{0}=\frac{\mu NI_{0}}{L'}$  

Divide this expression by the equation (1) by putting $L'=\frac{L}{4}$.

$B'_{0}=4B_{0$

Therefore, the option (iv) will work.

Consider the option (v) by using equation (1).

$B'_{0}=\frac{\mu NI_{0}}{L}$  

Divide this expression by the equation (1) by putting $L'=\frac{L}{2}$ and N=2N.

$B'_{0}=8B_{0$

Therefore, the option (v) won't work.

Consider the option (vi) by using equation (1).

$B_{0}=\frac{\mu N'I'_{0}}{L}$  

Divide the above expression by the equation (1) by putting $I'_{0}=2I_{0}$  and N=2N .

$B'_{0}=4B_{0$

Therefore, the option (vi) will work.

Therefore, the options (ii),(iv) and (vi) will work.