153849
Two long parallel wires are at a distance of $1 \mathrm{~m}$. If both of them carry $1 \mathrm{~A}$ of current. The force of attracting per unit length between the two wires is
1 $2 \times 10^{-7} \mathrm{~N} / \mathrm{m}$
2 $2 \times 10^{-8} \mathrm{~N} / \mathrm{m}$
3 $5 \times 10^{-8} \mathrm{~N} / \mathrm{m}$
4 $10^{-7} \mathrm{~N} / \mathrm{m}$
Explanation:
A Given, $I_{1}$ and $I_{2}$ are current that are flowing in the wires $\therefore \quad \mathrm{I}_{1}=\mathrm{I}_{2}=1 \mathrm{~A}$ Distance between wires $(\mathrm{d})=1 \mathrm{~m}$ We have, $\mathrm{F}=\frac{\mu_{\mathrm{o}}}{2 \pi} \frac{\mathrm{I}_{1} \mathrm{I}_{2}}{\mathrm{~d}} \times l$ $\frac{\mathrm{F}}{l}=\frac{\mu_{\mathrm{o}}}{2 \pi} \frac{\left(\mathrm{I}_{1} \mathrm{I}_{2}\right)}{\mathrm{d}}$ Where $\frac{\mathrm{F}}{l}$ denotes force per unit length $\frac{\mathrm{F}}{l}=\frac{4 \pi \times 10^{-7}}{2 \pi} \times \frac{(1)^{2}}{1}$ $\frac{\mathrm{F}}{l}=2 \times 10^{-7} \mathrm{~N} / \mathrm{m}$
CBESE AIPMT 1998
Moving Charges & Magnetism
153851
The magnetic force acting on a charged particle of charge $-2 \mu \mathrm{C}$ in a magnetic field of 2 $T$ acting in y-direction, when the particle velocity is $(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}) \times 10^{6} \mathrm{~ms}^{-1}$ is
153852
A charged particle is moving along a magnetic field line. The magnetic force on the particle is
1 along its velocity
2 opposite to its velocity
3 perpendicular to its velocity
4 zero
Explanation:
D As the charge is moving along the magnetic field line then direction of velocity and magnetic field will in the same direction. Therefore, the magnetic force on the particle will be zero. $\mathrm{F}$ $=\mathrm{qvB} \sin \theta$ $\theta=0^{\circ}$ $\therefore \quad \mathrm{F} =0$
DCE-2009
Moving Charges & Magnetism
153854
If a bar magnet of moment $\mu$ is suspended in a uniform magnetic field $B$ and its is given an angular deflection $\theta$ w.r.t. its equilibrium position, the restoring torque on magnet is
1 $\mu \mathrm{B} \sin \theta$
2 $\mu \mathrm{B} \cos \theta$
3 $\mu \mathrm{B} \tan \theta$
4 $\mu^{2} B^{2} \sin \theta \cos \theta$
Explanation:
A Given, Bar of magnet of moment $=\mu$ Where, $\mu=\mathrm{m}(2 l)$ Uniform magnetic field $=\mathrm{B}$ Angular deflection $=\theta$ Restoring torque $=\left(\tau_{\mathrm{R}}\right)=$ ? Now, as per conditions given in question We can write- $\because \quad \tau_{\mathrm{R}}=\tau_{\text {north }}+\tau_{\text {south }}$ $=\mathrm{m} / \mathrm{B} \sin \theta+\mathrm{m} / \mathrm{B} \sin \theta$ $=2 \mathrm{~m} / \mathrm{B} \sin \theta$ $\tau_{\mathrm{R}}=\mu \mathrm{B} \sin \theta$
153849
Two long parallel wires are at a distance of $1 \mathrm{~m}$. If both of them carry $1 \mathrm{~A}$ of current. The force of attracting per unit length between the two wires is
1 $2 \times 10^{-7} \mathrm{~N} / \mathrm{m}$
2 $2 \times 10^{-8} \mathrm{~N} / \mathrm{m}$
3 $5 \times 10^{-8} \mathrm{~N} / \mathrm{m}$
4 $10^{-7} \mathrm{~N} / \mathrm{m}$
Explanation:
A Given, $I_{1}$ and $I_{2}$ are current that are flowing in the wires $\therefore \quad \mathrm{I}_{1}=\mathrm{I}_{2}=1 \mathrm{~A}$ Distance between wires $(\mathrm{d})=1 \mathrm{~m}$ We have, $\mathrm{F}=\frac{\mu_{\mathrm{o}}}{2 \pi} \frac{\mathrm{I}_{1} \mathrm{I}_{2}}{\mathrm{~d}} \times l$ $\frac{\mathrm{F}}{l}=\frac{\mu_{\mathrm{o}}}{2 \pi} \frac{\left(\mathrm{I}_{1} \mathrm{I}_{2}\right)}{\mathrm{d}}$ Where $\frac{\mathrm{F}}{l}$ denotes force per unit length $\frac{\mathrm{F}}{l}=\frac{4 \pi \times 10^{-7}}{2 \pi} \times \frac{(1)^{2}}{1}$ $\frac{\mathrm{F}}{l}=2 \times 10^{-7} \mathrm{~N} / \mathrm{m}$
CBESE AIPMT 1998
Moving Charges & Magnetism
153851
The magnetic force acting on a charged particle of charge $-2 \mu \mathrm{C}$ in a magnetic field of 2 $T$ acting in y-direction, when the particle velocity is $(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}) \times 10^{6} \mathrm{~ms}^{-1}$ is
153852
A charged particle is moving along a magnetic field line. The magnetic force on the particle is
1 along its velocity
2 opposite to its velocity
3 perpendicular to its velocity
4 zero
Explanation:
D As the charge is moving along the magnetic field line then direction of velocity and magnetic field will in the same direction. Therefore, the magnetic force on the particle will be zero. $\mathrm{F}$ $=\mathrm{qvB} \sin \theta$ $\theta=0^{\circ}$ $\therefore \quad \mathrm{F} =0$
DCE-2009
Moving Charges & Magnetism
153854
If a bar magnet of moment $\mu$ is suspended in a uniform magnetic field $B$ and its is given an angular deflection $\theta$ w.r.t. its equilibrium position, the restoring torque on magnet is
1 $\mu \mathrm{B} \sin \theta$
2 $\mu \mathrm{B} \cos \theta$
3 $\mu \mathrm{B} \tan \theta$
4 $\mu^{2} B^{2} \sin \theta \cos \theta$
Explanation:
A Given, Bar of magnet of moment $=\mu$ Where, $\mu=\mathrm{m}(2 l)$ Uniform magnetic field $=\mathrm{B}$ Angular deflection $=\theta$ Restoring torque $=\left(\tau_{\mathrm{R}}\right)=$ ? Now, as per conditions given in question We can write- $\because \quad \tau_{\mathrm{R}}=\tau_{\text {north }}+\tau_{\text {south }}$ $=\mathrm{m} / \mathrm{B} \sin \theta+\mathrm{m} / \mathrm{B} \sin \theta$ $=2 \mathrm{~m} / \mathrm{B} \sin \theta$ $\tau_{\mathrm{R}}=\mu \mathrm{B} \sin \theta$
153849
Two long parallel wires are at a distance of $1 \mathrm{~m}$. If both of them carry $1 \mathrm{~A}$ of current. The force of attracting per unit length between the two wires is
1 $2 \times 10^{-7} \mathrm{~N} / \mathrm{m}$
2 $2 \times 10^{-8} \mathrm{~N} / \mathrm{m}$
3 $5 \times 10^{-8} \mathrm{~N} / \mathrm{m}$
4 $10^{-7} \mathrm{~N} / \mathrm{m}$
Explanation:
A Given, $I_{1}$ and $I_{2}$ are current that are flowing in the wires $\therefore \quad \mathrm{I}_{1}=\mathrm{I}_{2}=1 \mathrm{~A}$ Distance between wires $(\mathrm{d})=1 \mathrm{~m}$ We have, $\mathrm{F}=\frac{\mu_{\mathrm{o}}}{2 \pi} \frac{\mathrm{I}_{1} \mathrm{I}_{2}}{\mathrm{~d}} \times l$ $\frac{\mathrm{F}}{l}=\frac{\mu_{\mathrm{o}}}{2 \pi} \frac{\left(\mathrm{I}_{1} \mathrm{I}_{2}\right)}{\mathrm{d}}$ Where $\frac{\mathrm{F}}{l}$ denotes force per unit length $\frac{\mathrm{F}}{l}=\frac{4 \pi \times 10^{-7}}{2 \pi} \times \frac{(1)^{2}}{1}$ $\frac{\mathrm{F}}{l}=2 \times 10^{-7} \mathrm{~N} / \mathrm{m}$
CBESE AIPMT 1998
Moving Charges & Magnetism
153851
The magnetic force acting on a charged particle of charge $-2 \mu \mathrm{C}$ in a magnetic field of 2 $T$ acting in y-direction, when the particle velocity is $(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}) \times 10^{6} \mathrm{~ms}^{-1}$ is
153852
A charged particle is moving along a magnetic field line. The magnetic force on the particle is
1 along its velocity
2 opposite to its velocity
3 perpendicular to its velocity
4 zero
Explanation:
D As the charge is moving along the magnetic field line then direction of velocity and magnetic field will in the same direction. Therefore, the magnetic force on the particle will be zero. $\mathrm{F}$ $=\mathrm{qvB} \sin \theta$ $\theta=0^{\circ}$ $\therefore \quad \mathrm{F} =0$
DCE-2009
Moving Charges & Magnetism
153854
If a bar magnet of moment $\mu$ is suspended in a uniform magnetic field $B$ and its is given an angular deflection $\theta$ w.r.t. its equilibrium position, the restoring torque on magnet is
1 $\mu \mathrm{B} \sin \theta$
2 $\mu \mathrm{B} \cos \theta$
3 $\mu \mathrm{B} \tan \theta$
4 $\mu^{2} B^{2} \sin \theta \cos \theta$
Explanation:
A Given, Bar of magnet of moment $=\mu$ Where, $\mu=\mathrm{m}(2 l)$ Uniform magnetic field $=\mathrm{B}$ Angular deflection $=\theta$ Restoring torque $=\left(\tau_{\mathrm{R}}\right)=$ ? Now, as per conditions given in question We can write- $\because \quad \tau_{\mathrm{R}}=\tau_{\text {north }}+\tau_{\text {south }}$ $=\mathrm{m} / \mathrm{B} \sin \theta+\mathrm{m} / \mathrm{B} \sin \theta$ $=2 \mathrm{~m} / \mathrm{B} \sin \theta$ $\tau_{\mathrm{R}}=\mu \mathrm{B} \sin \theta$
153849
Two long parallel wires are at a distance of $1 \mathrm{~m}$. If both of them carry $1 \mathrm{~A}$ of current. The force of attracting per unit length between the two wires is
1 $2 \times 10^{-7} \mathrm{~N} / \mathrm{m}$
2 $2 \times 10^{-8} \mathrm{~N} / \mathrm{m}$
3 $5 \times 10^{-8} \mathrm{~N} / \mathrm{m}$
4 $10^{-7} \mathrm{~N} / \mathrm{m}$
Explanation:
A Given, $I_{1}$ and $I_{2}$ are current that are flowing in the wires $\therefore \quad \mathrm{I}_{1}=\mathrm{I}_{2}=1 \mathrm{~A}$ Distance between wires $(\mathrm{d})=1 \mathrm{~m}$ We have, $\mathrm{F}=\frac{\mu_{\mathrm{o}}}{2 \pi} \frac{\mathrm{I}_{1} \mathrm{I}_{2}}{\mathrm{~d}} \times l$ $\frac{\mathrm{F}}{l}=\frac{\mu_{\mathrm{o}}}{2 \pi} \frac{\left(\mathrm{I}_{1} \mathrm{I}_{2}\right)}{\mathrm{d}}$ Where $\frac{\mathrm{F}}{l}$ denotes force per unit length $\frac{\mathrm{F}}{l}=\frac{4 \pi \times 10^{-7}}{2 \pi} \times \frac{(1)^{2}}{1}$ $\frac{\mathrm{F}}{l}=2 \times 10^{-7} \mathrm{~N} / \mathrm{m}$
CBESE AIPMT 1998
Moving Charges & Magnetism
153851
The magnetic force acting on a charged particle of charge $-2 \mu \mathrm{C}$ in a magnetic field of 2 $T$ acting in y-direction, when the particle velocity is $(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}) \times 10^{6} \mathrm{~ms}^{-1}$ is
153852
A charged particle is moving along a magnetic field line. The magnetic force on the particle is
1 along its velocity
2 opposite to its velocity
3 perpendicular to its velocity
4 zero
Explanation:
D As the charge is moving along the magnetic field line then direction of velocity and magnetic field will in the same direction. Therefore, the magnetic force on the particle will be zero. $\mathrm{F}$ $=\mathrm{qvB} \sin \theta$ $\theta=0^{\circ}$ $\therefore \quad \mathrm{F} =0$
DCE-2009
Moving Charges & Magnetism
153854
If a bar magnet of moment $\mu$ is suspended in a uniform magnetic field $B$ and its is given an angular deflection $\theta$ w.r.t. its equilibrium position, the restoring torque on magnet is
1 $\mu \mathrm{B} \sin \theta$
2 $\mu \mathrm{B} \cos \theta$
3 $\mu \mathrm{B} \tan \theta$
4 $\mu^{2} B^{2} \sin \theta \cos \theta$
Explanation:
A Given, Bar of magnet of moment $=\mu$ Where, $\mu=\mathrm{m}(2 l)$ Uniform magnetic field $=\mathrm{B}$ Angular deflection $=\theta$ Restoring torque $=\left(\tau_{\mathrm{R}}\right)=$ ? Now, as per conditions given in question We can write- $\because \quad \tau_{\mathrm{R}}=\tau_{\text {north }}+\tau_{\text {south }}$ $=\mathrm{m} / \mathrm{B} \sin \theta+\mathrm{m} / \mathrm{B} \sin \theta$ $=2 \mathrm{~m} / \mathrm{B} \sin \theta$ $\tau_{\mathrm{R}}=\mu \mathrm{B} \sin \theta$
