CBSE 2025, Set 3

Question

(b) (i) (I) Write Ampere's circuital law in mathematical form and explain the terms used.
(II) As the current carrying solenoid is made longer, the magnetic field produced outside it approaches zero. Why?
(III) A flexible loop of irregular shape carrying current when located in an external magnetic field, changes to a circular shape. Give reason.
(ii) A galvanometer of resistance $G$ is converted into a voltmeter to measure up to $V$ volts, by connecting a resistance $R_1$ in series with the coil. If $R_1$ is replaced by $R_2$, then it can only measure up to $\frac{V}{2}$ volt. Find the value of the resistance $R_3$ (in terms of $R_1$ and $R_2$) needed to convert it into a voltmeter that can read up to $2V$.

Edvaya · Accepted Answer

The correct answer is Subjective. **(i) (I) Ampere's Circuital Law:**
Mathematical form: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_e$
* $\vec{B}$ = Magnetic field vector
* $d\vec{l}$ = Infinitesimal line element vector along the closed loop (Amperian loop)
* $\mu_0$ = Permeability of free space
* $I_e$ = Total net current enclosed by the closed Amperian loop.

**(II) Magnetic field outside a long solenoid:**
As the length of the solenoid increases, it closely approximates an ideal infinite solenoid. It appears like a long cylindrical uniform metal sheet of current. The magnetic field lines outside cancel each other out, making the field approach zero.

**(III) Flexible loop changing shape:**
A current-carrying loop in an external magnetic field experiences forces that tend to expand it to maximize the magnetic flux passing through it. For a given perimeter, a circle encloses the maximum possible area. Hence, the loop adopts a circular shape to maximize flux.

**(ii) Voltmeter Conversion:**
Let $I_g$ be the full-scale deflection current of the galvanometer.
When resistance $R_1$ is connected, the range is $V$:
$$V = I_g(G + R_1) \Rightarrow \frac{V}{I_g} = R_1 + G \quad 	ext{--- (1)}$$

When resistance $R_2$ is connected, the range is $\frac{V}{2}$:
$$\frac{V}{2} = I_g(G + R_2) \Rightarrow \frac{V}{2I_g} = R_2 + G \quad 	ext{--- (2)}$$

Subtracting (2) from (1):
$$\frac{V}{I_g} - \frac{V}{2I_g} = R_1 - R_2$$
$$\frac{V}{2I_g} = R_1 - R_2$$
Substituting this back into (2):
$$R_1 - R_2 = R_2 + G \Rightarrow G = R_1 - 2R_2$$

Now, to measure up to $2V$, let the required resistance be $R_3$:
$$2V = I_g(G + R_3) \Rightarrow \frac{2V}{I_g} = R_3 + G$$
$$R_3 = \frac{2V}{I_g} - G$$
From (1), we know $\frac{V}{I_g} = R_1 + G$. Therefore:
$$R_3 = 2(R_1 + G) - G = 2R_1 + 2G - G = 2R_1 + G$$
Substitute the value of $G$:
$$R_3 = 2R_1 + (R_1 - 2R_2)$$
$$\boxed{R_3 = 3R_1 - 2R_2}$$

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