Moving Charges and Magnetism Previous Year Questions — JEE Main Physics
42 past-year Moving Charges and Magnetism questions from JEE Main (Physics) — page 5.
Q39 (2024)
Two particles X and Y having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describes circular paths of radii $R_1$ and $R_2$ respectively. The mass ratio of X and Y is:
- $\left(\frac{R_2}{R_1}\right)^2$
- $\left(\frac{R_1}{R_2}\right)^2$
- $\left(\frac{R_1}{R_2}\right)$
- $\left(\frac{R_2}{R_1}\right)$
### Related Formula
The radius $R$ of the path of a charged particle moving perpendicular to a magnetic field $B$ is:
$$R = \frac{mv}{qB} = \frac{p}{qB}$$
In terms of kinetic energy $K$:
$$R = \frac{\sqrt{2mK}}{qB}$$
Since the particle is accelerated through potential $V$, kinetic energy $K = qV$:
$$R = \frac{\sqrt{2mqV}}{qB}$$
$$R = \frac{1}{B} \sqrt{\frac{2mV}{q}}$$
### Core Logic
For both particles X and Y, the following parameters are the same:
* Potential Difference, $V$
* Magnetic Field, $B$
* Charge, $q$
Therefore, we have the proportionality:
$$R \propto \sqrt{m} \implies R^2 \propto m$$
### Step 1: Calculate Mass Ratio
Using the proportionality relationship:
$$\frac{m_1}{m_2} = \left( \frac{R_1}{R_2} \right)^2$$
Thus, the mass ratio of X and Y is $\left(\frac{R_1}{R_2}\right)^2$.
### Pattern Recognition
Shortcut: Whenever charges and potential differences are equal, the radius of orbit in a magnetic field scales as $R \propto \sqrt{m}$. Squaring both sides yields $m \propto R^2$ instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q54 (2024)
A charge of $4.0\ \mu\text{C}$ is moving with a velocity of $4.0 \times 10^{6}\text{ ms}^{-1}$ along the positive y-axis under a magnetic field $\vec{B}$ of strength $(2\hat{k})\text{ T}$. The force acting on the charge is $x\hat{i}\text{ N}$. The value of $x$ is ________.
### Related Formula
The magnetic force on a moving charge is given by the Lorentz force equation:
$$\vec{F} = q (\vec{v} \times \vec{B})$$
### Core Logic
Given data:
* Charge, $q = 4.0\ \mu\text{C} = 4.0 \times 10^{-6}\text{ C}$
* Velocity vector, $\vec{v} = (4.0 \times 10^{6}\hat{j})\text{ ms}^{-1}$
* Magnetic field vector, $\vec{B} = 2\hat{k}\text{ T}$
### Step 1: Calculate the Force Vector
Substitute the values into the force equation:
$$\vec{F} = (4.0 \times 10^{-6}) \left[ (4.0 \times 10^{6}\hat{j}) \times (2\hat{k}) \right]$$
$$\vec{F} = 4.0 \times 10^{-6} \times 8.0 \times 10^{6} (\hat{j} \times \hat{k})$$
Since $\hat{j} \times \hat{k} = \hat{i}$:
$$\vec{F} = 32\hat{i}\text{ N}$$
Comparing this to $x\hat{i}\text{ N}$, we find:
$x = 32$
### Pattern Recognition
Cross-product check: $\hat{j} \times \hat{k} = \hat{i}$. The product of the scalar terms $(4.0 \times 10^{-6}) \times (4.0 \times 10^6) \times 2$ immediately simplifies to $4 \times 4 \times 2 = 32$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q1558 (2024)
Two long, straight wires carry equal currents in opposite directions as shown in the figure. The separation between the wires is $5.0\text{ cm}$. The magnitude of the magnetic field at a point $P$ midway between the wires is ______ $\mu\text{T}$. (Given : $\mu_{0} = 4\pi \times 10^{-7}\text{ T}\cdot\text{m}\cdot\text{A}^{-1}$, and each wire carries a current of $10\text{ A}$).
{{IMG1}}
### Related Formula
$$B = \frac{\mu_0 i}{2\pi r}$$
### Core Logic
Using the right-hand grip rule, both anti-parallel wire systems generate field arrays pointing in the exact same direction at the central midway coordinate. Hence, their field contributions add up directly:
$$B_{\text{net}} = 2 B_1 = 2 \left(\frac{\mu_0 i}{2\pi r}\right) = \frac{\mu_0 i}{\pi r}$$
Where $i = 10\text{ A}$, total distance $= 5\text{ cm} \implies r = 2.5\text{ cm} = 2.5 \times 10^{-2}\text{ m}$.
### Step 1: Compute numeric value
$$B_{\text{net}} = \frac{4\pi \times 10^{-7} \times 10}{\pi \times (2.5 \times 10^{-2})} = \frac{4 \times 10^{-6}}{2.5 \times 10^{-2}}$$
$$B_{\text{net}} = 1.6 \times 10^{-4}\text{ T} = 160 \times 10^{-6}\text{ T} = 160\ \mu\text{T}$$
### Pattern Recognition
Anti-parallel current pairs generate collaborative field additions inside their spatial boundary zone, rather than destructive structural cancellations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q34 (2024)
A proton moving with a constant velocity passes through a region of space without any change in its velocity.
If $\vec{E}$ and $\vec{B}$ represent the electric and magnetic fields respectively, then the region of space may have:
(A) $E = 0, B = 0$
(B) $E = 0, B \neq 0$
(C) $E \neq 0, B = 0$
(D) $E \neq 0, B \neq 0$
Choose the most appropriate answer from the options given below:
- $\text{(A), (B) and (C) only}$
- $\text{(A), (C) and (D) only}$
- $\text{(A), (B) and (D) only}$
- $\text{(B), (C) and (D) only}$
### Related Formula
$$\vec{F}_{\text{net}} = q\vec{E} + q(\vec{v} \times \vec{B})$$
### Core Logic
For velocity to remain constant, the net force must be zero:
$$q\vec{E} + q(\vec{v} \times \vec{B}) = 0$$
Let's evaluate the cases:
- Case (A): If $E=0$ and $B=0$, $\vec{F} = 0$. (Possible)
- Case (B): If $E=0$ and $B \neq 0$, the magnetic force is zero if $\vec{v}$ is parallel or antiparallel to $\vec{B}$ (i.e., $\vec{v} \times \vec{B} = 0$). (Possible)
- Case (C): If $E \neq 0$ and $B=0$, $\vec{F} = q\vec{E} \neq 0$, velocity must change. (Not possible)
- Case (D): If $E \neq 0$ and $B \neq 0$, the electric and magnetic forces can balance each other perfectly if $q\vec{E} = -q(\vec{v} \times \vec{B})$. (Possible)
### Step 1: Conclusion
Hence, statements (A), (B), and (D) represent valid situations where velocity can remain constant.
### Pattern Recognition
Velocity filter / velocity selector setups utilize crossed fields where electric fields perfectly balance magnetic components, an archetype of Case (D).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q1632 (2024)
A galvanometer having coil resistance $10 \, \Omega$ shows a full scale deflection for a current of $3 \mathrm{~mA}$. For it to measure a current of $8 \mathrm{~A}$, the value of the shunt should be:
- $3 \times 10^{-3} \, \Omega$
- $4.85 \times 10^{-3} \, \Omega$
- $3.75 \times 10^{-3} \, \Omega$
- $2.75 \times 10^{-3} \, \Omega$
### Related Formula
To convert a galvanometer into an ammeter, a small resistance called a shunt ($S$) is connected in parallel with the galvanometer ($G$):
$$I_g G = (I - I_g) S \implies S = \frac{I_g G}{I - I_g}$$
where,
$G$ = resistance of the galvanometer coil
$I_g$ = full-scale deflection current of the galvanometer
$I$ = total current to be measured
$S$ = shunt resistance
### Core Logic
Given values:
$$G = 10 \, \Omega$$
$$I_g = 3 \mathrm{~mA} = 3 \times 10^{-3} \mathrm{~A}$$
$$I = 8 \mathrm{~A}$$
### Step 1: Calculate Shunt Resistance
Substituting these values into the shunt formula:
$$S = \frac{(3 \times 10^{-3}) \times 10}{8 - 0.003}$$
$$S = \frac{0.03}{7.997}$$
$$S \approx 3.75 \times 10^{-3} \, \Omega$$
Therefore, the required value of the shunt resistance is $3.75 \times 10^{-3} \, \Omega$.
{{SOL_IMG1}}
### Pattern Recognition
Since $I \gg I_g$, the value $I - I_g$ in the denominator can be approximated directly as $I$ for a very accurate quick shortcut in choice selection: $S \approx \frac{I_g G}{I} = \frac{0.03}{8} = 3.75 \times 10^{-3} \, \Omega$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
