Four identical long solenoids A, B, C and D are connected to each other as shown in the figure. If t

Q: Why does the current divide equally among solenoids B, C, and D?

Because the problem states the solenoids are 'identical'. Identical solenoids made of the same wire will have the exact same electrical resistance. In a parallel circuit, current divides equally among branches with equal resistance.

Q: Does the length or radius of the solenoid affect the answer?

The formula for a long solenoid's magnetic field is $B = \mu_0 n i$, which depends on the turn density $n$ and current $i$. Since they are identical, $n$ is the same for all. The length and radius would only matter if they were different, as that would change their resistance and turn density.

Q: What if solenoid A had twice the number of turns as the others?

If solenoid A had twice the turn density ($2n$), its magnetic field would be $B_A = \mu_0 (2n) i$. The current would still divide equally among B, C, and D (if they are still identical to each other), so $B_C = \mu_0 n (i/3)$. The ratio would then be $B_A / B_C = 6$.

Q: Why is the magnetic field of B, C, and D not adding up to affect C?

The problem explicitly states to 'Assume that the magnetic field is confined within the volume of the respective solenoid.' This means we can ignore any magnetic interaction or flux leakage between the solenoids.

Q: Can we use Ampere's Law directly on the whole circuit?

Ampere's Law is used to derive the formula $B = \mu_0 n i$ for a single solenoid. For the circuit as a whole, you must apply Kirchhoff's Current Law to find the current in each individual solenoid first, and then apply the magnetic field formula to each one separately.

Question

Four identical long solenoids A, B, C and D are connected to each other as shown in the figure. If the magnetic field at the centre of A is 3T, the field at the centre of C would be (Assume that, the magnetic field is confined with in the volume of respective solenoid)

Options:

(A) 12T

(B) 6T

(C) 9T

(D) 1T

Sigma Percentile · Accepted Answer

Imagine you are tracing the path of an electric current as it flows through a circuit. In this problem, we have a fascinating setup: four identical long solenoids connected together. Solenoid A is in series with a parallel combination of solenoids B, C, and D.

### Analyzing the Setup
When the total current $i$ enters the circuit, it first passes entirely through solenoid A. After exiting A, the current reaches a junction where it must split into three separate branches containing solenoids B, C, and D. Because the problem states that all four solenoids are **identical**, we know they must have the exact same electrical resistance.

### The Master Equation
To understand what happens to the magnetic field, we need to recall the formula for the magnetic field at the center of a long solenoid. The magnetic field $B$ is given by the equation:

$$B = \mu_0 n i$$

where $\mu_0$ is the permeability of free space, $n$ is the number of turns per unit length, and $i$ is the current. Since all the solenoids are identical, their turn density $n$ is a constant. This reveals a beautiful, direct relationship: the magnetic field is directly proportional to the current flowing through the solenoid ($B \propto i$).

### Current Division
Now, let's look at the parallel branches. When the total current $i$ reaches the junction, it sees three identical paths. According to the laws of parallel circuits, the current will divide equally among branches with equal resistance. Therefore, the current flowing through solenoid C (and also B and D) will be exactly one-third of the total current:

$$i_C = \frac{i}{3}$$

### Final Calculation
Since the magnetic field is directly proportional to the current, the ratio of the magnetic fields in solenoids A and C will be equal to the ratio of their respective currents. We can set up a simple proportion:

$$\frac{B_A}{B_C} = \frac{i_A}{i_C}$$

Substituting the currents we found:

$$\frac{B_A}{B_C} = \frac{i}{\frac{i}{3}} = 3$$

We are given that the magnetic field at the center of solenoid A is $3	ext{ T}$. Plugging this into our equation:

$$\frac{3}{B_C} = 3$$

Solving for $B_C$, we find:

$$B_C = 1	ext{ T}$$

And there we have it! The magnetic field at the center of solenoid C is exactly $1	ext{ T}$. This problem beautifully demonstrates how electrical circuit principles like current division can directly impact magnetic phenomena.

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Visualized Solution

$C i r c u i t A na l y s i s$

$M a g n e t i c F i e l d F or m u l a$

$C u r r e n t D i v i s i o n$

$R a t i o S e t u p$

$F ina l C a l c u l a t i o n$

$T h e W a y F or w a r d$

The Sigma Insight: Ampere's Circuital Law

Analyzing the Setup When the total current $i$ enters the circuit, it first passes entirely through solenoid A

The Master Equation To understand what happens to the magnetic field, we need to recall the formula for the magnetic field at the center of a long solenoid

Current Division Now, let's look at the parallel branches

Final Calculation Since the magnetic field is directly proportional to the current, the ratio of the magnetic fields in solenoids A and C will be equal to the ratio of their respective currents

Similar Questions

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Circuit Analysis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Field Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Current Division

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward