Moving Charges and Magnetism Previous Year Questions — JEE Main Physics
42 past-year Moving Charges and Magnetism questions from JEE Main (Physics) — page 6.
Q1634 (2024)
The deflection in moving coil galvanometer falls from 25 divisions to 5 division when a shunt of $24 \, \Omega$ is applied. The resistance of galvanometer coil will be:
- $12 \, \Omega$
- $96 \, \Omega$
- $48 \, \Omega$
- $100 \, \Omega$
### Related Formula
For a shunted galvanometer, the potential difference across the galvanometer is equal to the potential difference across the shunt resistor:
$$I_g G = (I - I_g) S$$
### Core Logic
Let the current sensitivity per division be $x$.
* Initially, without the shunt, the total current $I$ gives a full scale deflection of 25 divisions:
$I = 25x$
{{SOL_IMG1}}
* When the shunt ($S = 24 \, \Omega$) is connected in parallel, the current passing through the galvanometer branch ($I_g$) corresponds to a deflection of 5 divisions:
$I_g = 5x$
{{SOL_IMG2}}
### Step 1: Find Current through the Shunt
The current remaining for the shunt branch is:
$$I - I_g = 25x - 5x = 20x$$
### Step 2: Equate Potential Drops
Applying the parallel branch condition:
$$5x \cdot G = 20x \cdot 24$$
Dividing both sides by $5x$:
$$G = 4 \times 24 = 96 \, \Omega$$
Therefore, the resistance of the galvanometer coil is $96 \, \Omega$.
### Pattern Recognition
The deflection is directly proportional to the branch current. A drop from 25 divisions to 5 divisions means only $\frac{5}{25} = \frac{1}{5}$ of the total current flows through the galvanometer, leaving $\frac{4}{5}$ to go through the shunt. Hence, the galvanometer resistance must be 4 times the shunt resistance.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q56 (2024)
The current of $5 \mathrm{~A}$ flows in a square loop of sides $1 \mathrm{~m}$ is placed in air. The magnetic field at the centre of the loop is $\mathrm{X}\sqrt{2} \times 10^{-7} \mathrm{T}$. The value of $\mathrm{X}$ is ________
### Related Formula
$$B = \frac{\mu_0 i}{4\pi d} (\sin\theta_1 + \sin\theta_2)$$
### Core Logic
For a square loop of side length $a$, the perpendicular distance from the center to any side is $d = \frac{a}{2}$.
The angles subtended by the ends of one side at the center are $\theta_1 = \theta_2 = 45^\circ$.
Since there are 4 identical sides contributing to the field in the same direction, the total magnetic field is $4$ times the field due to one side.
### Step 1: Calculate Field for One Side
$$B_1 = \frac{\mu_0 i}{4\pi (a/2)} (\sin 45^\circ + \sin 45^\circ)$$
$$B_1 = \frac{4\pi \times 10^{-7} \times 5}{4\pi \times 0.5} \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right)$$
$$B_1 = \frac{10^{-7} \times 5}{0.5} \times \frac{2}{\sqrt{2}} = 10 \times 10^{-7} \times \sqrt{2}$$
### Step 2: Total Magnetic Field
$$B_{\text{total}} = 4 \times B_1 = 4 \times 10\sqrt{2} \times 10^{-7} \mathrm{~T}$$
$$B_{\text{total}} = 40\sqrt{2} \times 10^{-7} \mathrm{~T}$$
Comparing with $X\sqrt{2} \times 10^{-7} \mathrm{T}$, we get $X = 40$.
### Pattern Recognition
The magnetic field at the center of an $n$-sided regular polygon carrying current $I$ and circumscribed by circle of radius $R$ is $B = \frac{\mu_0 n I}{2\pi R} \sin\left(\frac{\pi}{n}\right)$. Or simply use $B = 4 \times B_{\text{wire}}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q39 (2024)
Two insulated circular loop A and B radius 'a' carrying a current of 'I' in the anti clockwise direction as shown in figure. The magnitude of the magnetic induction at the centre will be :
{{IMG1}}
- $\frac{\sqrt{2} \mu_0 I}{a}$
- $\frac{\mu_0 I}{2 a}$
- $\frac{\mu_0 I}{\sqrt{2} a}$
- $\frac{2 \mu_0 I}{a}$
### Related Formula
$$B = \frac{\mu_0 I}{2R}$$
$$B_{\text{net}} = \sqrt{B_1^2 + B_2^2} \quad (\text{for mutually perpendicular fields})$$
### Core Logic
The two loops are mutually perpendicular to each other. The magnetic field at the center due to each loop acts along their respective normal axes, meaning the two resulting magnetic field vectors are orthogonal ($90^\circ$ apart).
{{SOL_IMG1}}
### Step 1: Calculate Individual Fields
Magnetic field at the center due to loop A:
$$B_A = \frac{\mu_0 I}{2a}$$
Magnetic field at the center due to loop B:
$$B_B = \frac{\mu_0 I}{2a}$$
### Step 2: Resultant Magnetic Field
Since $B_A$ and $B_B$ are perpendicular to each other:
$$B_{\text{net}} = \sqrt{B_A^2 + B_B^2}$$
$$B_{\text{net}} = \sqrt{\left(\frac{\mu_0 I}{2a}\right)^2 + \left(\frac{\mu_0 I}{2a}\right)^2}$$
$$B_{\text{net}} = \frac{\mu_0 I}{2a} \sqrt{2} = \frac{\sqrt{2} \mu_0 I}{2a} = \frac{\mu_0 I}{\sqrt{2} a}$$
### Pattern Recognition
When two identical physical sources generate orthogonal vector fields of magnitude $B$, the resultant is simply $\sqrt{2}B$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q51 (2024)
The horizontal component of earth's magnetic field at a place is $3.5 \times 10^{-5} \mathrm{~T}$. A very long straight conductor carrying current of $\sqrt{2} \mathrm{~A}$ in the direction from South east to North West is placed. The force per unit length experienced by the conductor is $\dots \dots \dots \times 10^{-6} \mathrm{~N/m}$.
### Related Formula
$$F = I L B \sin \theta$$
$$\frac{F}{L} = I B \sin \theta$$
### Core Logic
Earth's horizontal magnetic field ($B_H$) points strictly from South to North. The conductor is aligned from South-East to North-West. Thus, the angle $\theta$ between the current flow direction (SE to NW) and Earth's magnetic field (S to N) is exactly $45^\circ$.
### Step 1: Compute Force per Unit Length
$B_H = 3.5 \times 10^{-5} \mathrm{~T}$
$i = \sqrt{2} \mathrm{~A}$
$\theta = 45^\circ$
$$\frac{F}{\ell} = i B_H \sin \theta$$
$$\frac{F}{\ell} = \sqrt{2} \times (3.5 \times 10^{-5}) \times \frac{1}{\sqrt{2}}$$
$$\frac{F}{\ell} = 3.5 \times 10^{-5} \mathrm{~N/m}$$
$$\frac{F}{\ell} = 35 \times 10^{-6} \mathrm{~N/m}$$
### Step 2: Extract Value
The required multiplier for $10^{-6}$ is 35.
### Pattern Recognition
Geographical alignments often yield clean angles like $45^\circ$ or $90^\circ$. S-N field intersecting SE-NW current creates a $45^\circ$ angular intersection, perfectly canceling the $\sqrt{2}$ amp current.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Class 12 Physics: Magnetism and Matter
Q32 (2024)
A uniform magnetic field of $2 \times 10^{-3} \text{ T}$ acts along positive Y-direction. A rectangular loop of sides $20 \text{ cm}$ and $10 \text{ cm}$ with current of $5 \text{ A}$ is Y-Z plane. The current is in anticlockwise sense with reference to negative X axis. Magnitude and direction of the torque is :
- $2 \times 10^{-4} \text{ N-m}$ along positive Z-direction
- $2 \times 10^{-4} \text{ N-m}$ along negative Z-direction
- $2 \times 10^{-4} \text{ N-m}$ along positive X-direction
- $2 \times 10^{-4} \text{ N-m}$ along positive Y-direction
### Related Formula
$$\vec{M} = i\vec{A}$$
$$\vec{\tau} = \vec{M} \times \vec{B}$$
### Core Logic
The area vector $\vec{A}$ is perpendicular to the plane of the loop. According to the right-hand rule, an anticlockwise current observed from the negative X-axis means the area vector points in the negative X-direction.
{{SOL_IMG1}}
{{SOL_IMG2}}
### Step 1: Calculating Magnetic Moment
Area of the loop, $A = 0.2 \text{ m} \times 0.1 \text{ m} = 0.02 \text{ m}^2$
Magnetic moment vector:
$$\vec{M} = i \vec{A}$$
$$\vec{M} = 5 \times (0.2) \times (0.1) (-\hat{i})$$
$$\vec{M} = 0.1(-\hat{i}) \text{ A m}^2$$
### Step 2: Calculating Torque
The magnetic field is $\vec{B} = 2 \times 10^{-3} \hat{j} \text{ T}$.
Torque on the loop:
$$\vec{\tau} = \vec{M} \times \vec{B}$$
$$\vec{\tau} = 0.1(-\hat{i}) \times (2 \times 10^{-3} \hat{j})$$
$$\vec{\tau} = 2 \times 10^{-4} (-\hat{k}) \text{ N-m}$$
### Step 3: Final Direction
The $-\hat{k}$ direction corresponds to the negative Z-direction. Magnitude is $2 \times 10^{-4} \text{ N-m}$.
### Pattern Recognition
Identify the plane of the loop to get the possible normal vectors. Since the loop is in the Y-Z plane, $\vec{A}$ must be $\pm\hat{i}$. 'Anticlockwise from -X' fixes it as $-\hat{i}$. Simply apply cross product $-\hat{i} \times \hat{j} = -\hat{k}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
