NEET Moving Charges and Magnetism Previous Year Questions with Solutions

Semicircular segments ABC and ADC are connected in parallel between nodes A and C:

• • Given R_{ABC} = \frac{1}{2} R_{ADC} \implies I_{ABC} = 2 I_{ADC}.
• • Kirchhoff's Current Law at node A: I_{ABC} + I_{ADC} = I_0 \implies 2 I_{ADC} + I_{ADC} = 3 I_{ADC} = I_0.
• • Therefore: I_{ADC} = \frac{1}{3} I_0, \quad I_{ABC} = \frac{2}{3} I_0.

Use the Biot-Savart law for a circular arc spanning angle pi:

• • Branch ABC (current flows A -> B -> C, clockwise): Magnetic field B_{ABC} points perpendicularly INTO the page (\otimes): B_{ABC} = \frac{\mu_0 (2/3 I_0)}{4r} = \frac{2\mu_0 I_0}{12r}.
• • Branch ADC (current flows A -> D -> C, counter-clockwise): Magnetic field B_{ADC} points perpendicularly OUT OF the page (\odot): B_{ADC} = \frac{\mu_0 (1/3 I_0)}{4r} = \frac{\mu_0 I_0}{12r}.

Subtract the smaller field from the larger field to obtain net magnitude:

• • The net magnetic field points into the page with magnitude mu_0 * I_0 / (12r).
• • Matches Option (1) exactly.

According to Ampere's force law, two parallel conductors carrying currents in the same direction attract each other. The magnetic force per unit length exerted on wire A by fixed wire B is:

• • Because wire B is situated vertically above wire A, this attractive magnetic force acts vertically UPWARDS on wire A.

Two vertical forces act on wire A resting on the insulated floor:

• • N is the normal reaction from the floor per unit length.
• • Downwards gravitational force per unit length is lambda * g.

For wire A not to rise from the floor, normal reaction must be non-negative (N >= 0):

• • The minimum permissible height is h_{min} = \frac{\mu_0 I^2}{\pi \lambda g}.
• • Matches Option (2) exactly.

For the electron to travel without any deflection, the net Lorentz force acting on it must be zero:

• • Because v and B are mutually perpendicular, the vector (v × B) is perpendicular to both v and B.
• • Therefore, E must be strictly perpendicular to B.

Equating the magnitudes of electric and magnetic forces gives the electric field magnitude required:

• • Thus, E ⟂ B with magnitude 27 × 10² V/m (2700 V/m).

The magnetic dipole moment M of a planar loop of area A carrying current I is defined as:

• • M is directly proportional to the enclosed area.

For two circular coils carrying the same current I = 2 A with radii r₁ and r₂:

• • Thus the ratio of their respective magnetic moments is 1:4.

The magnitude of the magnetic field B at the center of a flat circular coil consisting of N closely wound turns carrying steady current I is given by the Biot-Savart law:

Substitute N = 100, I = 7 A, R = 10 cm = 0.1 m, and μ₀ = 4π × 10⁻⁷ T m/A:

Using π = 22/7 enables exact cancellation with the current of 7 A:

Dividing 88 by 0.2 gives 440:

A straight wire of length vector L⃗ carrying a steady current I placed in a uniform magnetic field B⃗ experiences a total magnetic force given by the vector cross product:

The wire carries current I along the positive x-axis, so its length displacement vector is directed entirely along î:

Using the cyclic right-handed rules for unit vectors (î × î = 0, î × ĵ = k̂, î × k̂ = -ĵ):

The magnitude of the resultant force vector is:

By the Biot-Savart Law, a circular loop of radius R carrying current i produces B = μ₀i / 2R at its center. A semicircle subtends an angle of π radians at center P, producing exactly half this value:

The top wire extends from x = -∞ to x = A at a perpendicular distance R above point P. By the standard Biot-Savart formula for a semi-infinite straight wire at an endpoint perpendicular distance R:

The bottom wire carries current i to the left from x = B to x = -∞ at a perpendicular distance R below point P. Applying the right-hand grip rule, its field at point P also points INTO the page:

The arc field points out of the page (⊙), while the sum of both straight wire fields points into the page (⊗). Subtracting the opposing magnitudes:

Since π ≈ 3.1416, 2/π ≈ 0.637 < 1, the bracket [1 - 2/π] is strictly positive (+0.363). Therefore, the outward arc field dominates, and the net magnetic field points AWAY FROM (out of) the page.

Using Ampere's circuital law / Biot-Savart law, the magnetic field at a perpendicular distance $r = 20\text{ cm} = 0.2\text{ m}$ from a long wire carrying current $I = 5\text{ A}$ is:

By the right-hand thumb rule, for current flowing from P to Q (towards the right), the magnetic field above the wire points out of the page ($\hat{k}$). The electron moves parallel to the wire ($\hat{i}$). Therefore, $\vec{v} \perp \vec{B}$, so $\theta = 90^\circ$ and $\sin 90^\circ = 1$.

The magnitude of the magnetic Lorentz force on the electron ($q = 1.6 \times 10^{-19}\text{ C}$) is:

Consider an Amperian loop of radius $r \le R$ concentric with the cable axis. Because current $I$ is uniformly distributed over the cross-sectional area $\pi R^2$, the current enclosed by the loop is:

By Ampere's circuital law $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$:

For an Amperian loop of radius $r \ge R$, the entire current $I$ is enclosed:

The field starts at 0 at the center ($r = 0$), rises linearly with distance $r$ up to the surface $r = R$, where it reaches a peak value, and then decreases asymptotically as $1/r$ for $r > R$. This corresponds uniquely to Graph (3).

For charge $q = 1$, the magnetic force is $\vec{F} = \vec{v} \times \vec{B}$. Given $\vec{v} = 2\hat{i} + 4\hat{j} + 6\hat{k}$ and $\vec{B} = B\hat{i} + B\hat{j} + B_0\hat{k}$:

Expand along the top row:

We are given $\vec{F} = 4\hat{i} - 20\hat{j} + 12\hat{k}$: 1. From the $\hat{k}$-component: $$-2B = 12 \implies B = -6$$ 2. From the $\hat{i}$-component: $$4 B_0 - 6 B = 4 \implies 4 B_0 - 6(-6) = 4 \implies 4 B_0 + 36 = 4 \implies 4 B_0 = -32 \implies B_0 = -8$$ 3. Verify the $\hat{j}$-component: $$-(2 B_0 - 6 B) = -[2(-8) - 6(-6)] = -[-16 + 36] = -20 \quad (\text{Matches } F_y = -20)$

Substitute $B = -6$ and $B_0 = -8$ into $\vec{B} = B\hat{i} + B\hat{j} + B_0\hat{k}$:

For a flat coil with $N$ turns, carrying current $I$ and enclosing cross-sectional area $A$, the magnitude of the magnetic dipole moment is:

The perimeter of one triangular loop of side $a$ is $P_1 = 3a$. With total length $L = 12a$: - Number of turns: $N_1 = \frac{L}{3a} = \frac{12a}{3a} = 4$ - Area of one equilateral triangle: $A_1 = \frac{\sqrt{3}}{4} a^2$

The perimeter of one square loop of side $a$ is $P_2 = 4a$. With total length $L = 12a$: - Number of turns: $N_2 = \frac{L}{4a} = \frac{12a}{4a} = 3$ - Area of one square: $A_2 = a^2$

By Ampere's circuital law, the magnetic field near the center inside a long current-carrying solenoid is uniform and given by: $$B = \mu_0 n I$$, where $$n$$ is the number of turns per unit length.

- Total turns: $$N = 100$$ - Length: $$L = 50\text{ cm} = 0.5\text{ m}$$ $$n = \frac{N}{L} = \frac{100}{0.5} = 200\text{ turns/m}$$.

$$B = \mu_0 n I = (4\pi \times 10^{-7}\text{ T m A}^{-1}) \times (200\text{ m}^{-1}) \times (2.5\text{ A})$$ $$B = 4\pi \times 10^{-7} \times 500 = 2000\pi \times 10^{-7} = 2\pi \times 10^{-4}\text{ T}$$ Substitute $$\pi \approx 3.14$$: $$B = 2 \times 3.14 \times 10^{-4} = 6.28 \times 10^{-4}\text{ T}$$.

Matches Option 1.