Force on a Current Carrying Conductor Questions in English

(B) Given: Mass of the wire, $m = 0.2 \,kg$. Length of the wire, $l = 1.5 \,m$. Current, $I = 2 \,A$. Acceleration due to gravity, $g = 10 \,m/s^2$.
The magnetic force $F_m$ acting on a current-carrying wire in a uniform magnetic field is given by $F_m = I l B \sin \theta$. Since the magnetic field is perpendicular to the wire, $\theta = 90^{\circ}$, so $F_m = I l B$.
For the wire to be suspended in mid-air, the upward magnetic force must balance the downward gravitational force (weight) of the wire.
$F_m = w$
$I l B = m g$
Rearranging to solve for $B$:
$B = \frac{m g}{I l}$
$B = \frac{0.2 \times 10}{2 \times 1.5}$
$B = \frac{2}{3} \approx 0.67 \,T$
Therefore, the magnitude of the magnetic field is $0.67 \,T$.

(C) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L} \times \vec{B})$.
Here,the length vector is $\vec{L} = L \hat{i}$.
Given $\vec{F} = I B_0 L(\hat{k} - \hat{j})$.
Substituting these into the formula: $I B_0 L(\hat{k} - \hat{j}) = I(L \hat{i} \times \vec{B})$.
Dividing by $I L$,we get: $B_0(\hat{k} - \hat{j}) = \hat{i} \times \vec{B}$.
Let $\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$.
Then $\hat{i} \times (B_x \hat{i} + B_y \hat{j} + B_z \hat{k}) = B_y \hat{k} - B_z \hat{j}$.
Comparing this with $B_0(\hat{k} - \hat{j})$,we get $B_y = B_0$ and $B_z = B_0$.
Since the cross product with $\hat{i}$ eliminates any component along $\hat{i}$,$B_x$ can be any value,but looking at the options,$B_x = B_0$ satisfies the condition.
Thus,$\vec{B} = B_0(\hat{i} + \hat{j} + \hat{k})$.

(C) The magnetic force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = I (\vec{L}_{eff} \times \vec{B})$,where $\vec{L}_{eff}$ is the effective displacement vector from the starting point to the end point of the wire.
For the given curve $y = A \sin \left(\frac{2 \pi}{\lambda} x\right)$,the starting point is at $x=0$,which gives $y=0$. The end point is at $x=\lambda$,which gives $y = A \sin(2\pi) = 0$.
Thus,the effective displacement vector $\vec{L}_{eff}$ is the vector from $(0, 0)$ to $(\lambda, 0)$,which is $\vec{L}_{eff} = \lambda \hat{i}$.
The magnetic field is in the $z$-direction,so $\vec{B} = B \hat{k}$.
The magnetic force is $\vec{F} = I (\lambda \hat{i} \times B \hat{k}) = I \lambda B (\hat{i} \times \hat{k}) = -I \lambda B \hat{j}$.
The magnitude of the magnetic force is $|\vec{F}| = I \lambda B$.

(C) The magnetic force on a small current element $I d\vec{l}$ is given by $d\vec{F} = I (d\vec{l} \times \vec{B})$.
Here,the current element is $d\vec{l} = dx \hat{i}$ and the magnetic field is $\vec{B} = B_0 \left( 2 - \frac{x}{a} \right) \hat{k}$.
Thus,$d\vec{F} = I (dx \hat{i}) \times B_0 \left( 2 - \frac{x}{a} \right) \hat{k} = I B_0 \left( 2 - \frac{x}{a} \right) dx (\hat{i} \times \hat{k}) = I B_0 \left( 2 - \frac{x}{a} \right) dx (-\hat{j})$.
To find the total force,we integrate from $x = a$ to $x = 2a$:
$\vec{F} = \int_{a}^{2a} I B_0 \left( 2 - \frac{x}{a} \right) dx (-\hat{j})$
$\vec{F} = -I B_0 \hat{j} \int_{a}^{2a} \left( 2 - \frac{x}{a} \right) dx$
$\vec{F} = -I B_0 \hat{j} \left[ 2x - \frac{x^2}{2a} \right]_{a}^{2a}$
$\vec{F} = -I B_0 \hat{j} \left[ \left( 2(2a) - \frac{(2a)^2}{2a} \right) - \left( 2(a) - \frac{a^2}{2a} \right) \right]$
$\vec{F} = -I B_0 \hat{j} \left[ (4a - 2a) - (2a - 0.5a) \right]$
$\vec{F} = -I B_0 \hat{j} [ 2a - 1.5a ] = -I B_0 \left( \frac{a}{2} \right) \hat{j}$.
Comparing this with the given force $\vec{F} = I B_0 \left( \frac{ka}{2} \right) \hat{j}$,we get $k = -1$.

(C) The magnetic force $F_B$ acting on a current-carrying wire in a magnetic field is given by $F_B = I \vec{L}_{eff} \times \vec{B}$.
For a semi-circular wire of radius $r$,the effective length $\vec{L}_{eff}$ is the straight-line distance between the two ends,which is the diameter $2r$.
Therefore,the magnitude of the magnetic force is $F_B = I(2r)B = 2IrB$.
This total upward magnetic force is shared equally by the two identical springs $X$ and $Y$ attached to the ends of the wire.
Let $F$ be the force acting on each spring. Then,$2F = F_B$.
Substituting the value of $F_B$,we get $2F = 2IrB$.
Thus,the force on each spring is $F = I r B$.

(A) The magnetic force on a current-carrying conductor in a uniform magnetic field is given by $\overrightarrow{F} = I(\overrightarrow{L} \times \overrightarrow{B})$,where $\overrightarrow{L}$ is the displacement vector from the starting point to the end point of the conductor.
For any closed loop,the starting point and the end point are the same,which means the total displacement vector $\overrightarrow{L}$ is zero.
Therefore,the net magnetic force $\overrightarrow{F}$ on any closed current loop placed in a uniform magnetic field is always zero.

(C) The force $\vec{F}$ on a current-carrying wire is given by $\vec{F} = I(\vec{L} \times \vec{B})$.
Here,$I = 1 \ A$,$\vec{L} = 0.5 \hat{i} \ m$,and $\vec{B} = (2\hat{i} + 4\hat{j}) \ T$.
Substituting the values:
$\vec{F} = 1 \times (0.5 \hat{i} \times (2\hat{i} + 4\hat{j}))$
$\vec{F} = 0.5 \times 2(\hat{i} \times \hat{i}) + 0.5 \times 4(\hat{i} \times \hat{j})$
Since $\hat{i} \times \hat{i} = 0$ and $\hat{i} \times \hat{j} = \hat{k}$:
$\vec{F} = 0 + 2\hat{k} = 2\hat{k} \ N$.
The magnitude of the force is $|\vec{F}| = 2 \ N$,and the direction is along the positive $z$-axis.

(D) The force $F$ on a current-carrying wire in a magnetic field is given by the formula $F = I L B \sin \theta$.
Given values are:
Current $I = 10 \ A$
Length $L = 2 \ m$
Magnetic field $B = 0.15 \ T$
Angle $\theta = 45^{\circ}$
Substituting these values into the formula:
$F = 10 \times 2 \times 0.15 \times \sin 45^{\circ}$
$F = 3 \times \frac{1}{\sqrt{2}}$
$F = \frac{3}{\sqrt{2}} \ N$.

(B) The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $d$ is given by $f = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
For wire $Q$ (current $I_Q = 1 \ A$):
$1$. Force due to wire $P$ ($I_P = 3 \ A$,$d_1 = 3 \ cm = 0.03 \ m$): Since currents are in opposite directions,the force is repulsive (away from $P$,i.e.,towards $R$).
$F_{QP} = \frac{\mu_0 I_Q I_P}{2 \pi d_1} \ell = (2 \times 10^{-7}) \times \frac{1 \times 3}{0.03} \times 0.15 = 2 \times 10^{-7} \times 100 \times 0.15 = 3 \times 10^{-6} \ N$ (towards $R$).
$2$. Force due to wire $R$ ($I_R = 2 \ A$,$d_2 = 2 \ cm = 0.02 \ m$): Since currents are in the same direction,the force is attractive (towards $R$).
$F_{QR} = \frac{\mu_0 I_Q I_R}{2 \pi d_2} \ell = (2 \times 10^{-7}) \times \frac{1 \times 2}{0.02} \times 0.15 = 2 \times 10^{-7} \times 100 \times 0.15 = 3 \times 10^{-6} \ N$ (towards $R$).
Total force $F_{net} = F_{QP} + F_{QR} = 3 \times 10^{-6} + 3 \times 10^{-6} = 6 \times 10^{-6} \ N$ towards $R$.

(A) The force $\vec{F}$ on a conductor of length $L$ carrying current $I$ in a magnetic field $\vec{B}$ is given by $\vec{F} = I(\vec{L} \times \vec{B})$.
The magnitude of the force per unit length $f$ is given by $f = I B \sin\theta$,where $\theta$ is the angle between the direction of the current and the magnetic field.
In this case,the magnetic field $\vec{B}$ is directed from South to North,and the current $I$ is also flowing from South to North.
Therefore,the angle $\theta$ between the current and the magnetic field is $0^\circ$.
Since $\sin(0^\circ) = 0$,the force per unit length $f = I \times B \times \sin(0^\circ) = 0 \text{ N/m}$.