Moving Charges and Magnetism

The magnetic moment M of a current-carrying loop is given by:M=IA,
where I is the current and A is the area of the loop. The area of a circular loop is:A=π$r^2$.
The circumference of the loop is 2πr=4.4m,

so:r=$\frac{4.4}{2\pi }$=0.7m.

Now, calculating the area:A=π($0.7^2$)=1.54
Thus, the magnetic moment is:M=1.0×1.54=1.54A$m^2$.

$B = \frac{{\mu }_{0}i}{2\pi r}$

from Right hand thumb rule, direction of B is North.

$B = \frac{{\mu }_{0}i}{2\pi r}$

from Right hand thumb rule, direction of B is North.

When the electric field is switched off, the magnetic force provides the centripetal force required for circular motion. The magnetic force acting on the proton is given by:

${\mathrm{F}}_{\mathrm{m}} = \mathrm{qvB} \sin 90° = \mathrm{qvB}$ ..... (1)

The centripetal force is given by

$\mathrm{Fc} = \frac{{\mathrm{mv}}^2}{\mathrm{r}}$ .... (2)

By equating equation (1) and (2), we get

$$\begin{gathered} \mathrm{qvB} =\frac{{\mathrm{mv}}^2}{\mathrm{r}} \\ \mathrm{B} = \frac{\mathrm{m} \mathrm{v}}{\mathrm{r} \mathrm{q}} \\ \mathrm{B} =\frac{1.67\times {10}^{-27}\times 2\times {10}^8}{1.6\times {10}^{-5}\times 1.6\times {10}^{-19}} \\ \mathrm{B} = 1.3\times {10}^5 \mathrm{T} \end{gathered}$$

\begin{aligned}
& \text{Given: Galvanometer resistance } = G\ \Omega, \quad \text{Full-scale current } = I\ \mathrm{A}, \\[3pt]
& \text{Current through galvanometer } = 0.1\% \text{ of } I = \frac{0.1}{100}I = \frac{I}{1000}. \\[6pt]
& \text{Let shunt resistance } = S. \text{ Then the same potential difference acts across } G \text{ and } S. \\[3pt]
& \text{So, } I_g G = I_s S, \quad \text{where } I_s = I - I_g = I - \frac{I}{1000} = \frac{999I}{1000}. \\[4pt]
& \Rightarrow S = \frac{I_g G}{I_s} = \frac{\frac{I}{1000} G}{\frac{999I}{1000}} = \frac{G}{999}. \\[6pt]
& \text{The equivalent resistance of the ammeter is } R_A = G \parallel S \\[4pt]
& R_A = \frac{G \times S}{G + S} = \frac{G \times \frac{G}{999}}{G + \frac{G}{999}} = \frac{G^2 / 999}{G(1 + 1/999)} \\[4pt]
& R_A = \frac{G / 999}{1 + 1/999} = \frac{G / 999}{1000/999} = \frac{G}{1000}. \\[6pt]
& \boxed{R_A = \frac{G}{1000}} \\[4pt]
& \text{Hence, the resistance of the ammeter is } \boxed{\tfrac{G}{1000}\ \Omega.}
\end{aligned}

$B = \frac{{\mu }_{0}i}{2\pi r}$

from Right hand thumb rule, direction of B is North.

\begin{aligned}
& \text{Given: Galvanometer resistance } = G\ \Omega, \quad \text{Full-scale current } = I\ \mathrm{A}, \\[3pt]
& \text{Current through galvanometer } = 0.1\% \text{ of } I = \frac{0.1}{100}I = \frac{I}{1000}. \\[6pt]
& \text{Let shunt resistance } = S. \text{ Then the same potential difference acts across } G \text{ and } S. \\[3pt]
& \text{So, } I_g G = I_s S, \quad \text{where } I_s = I - I_g = I - \frac{I}{1000} = \frac{999I}{1000}. \\[4pt]
& \Rightarrow S = \frac{I_g G}{I_s} = \frac{\frac{I}{1000} G}{\frac{999I}{1000}} = \frac{G}{999}. \\[6pt]
& \text{The equivalent resistance of the ammeter is } R_A = G \parallel S \\[4pt]
& R_A = \frac{G \times S}{G + S} = \frac{G \times \frac{G}{999}}{G + \frac{G}{999}} = \frac{G^2 / 999}{G(1 + 1/999)} \\[4pt]
& R_A = \frac{G / 999}{1 + 1/999} = \frac{G / 999}{1000/999} = \frac{G}{1000}. \\[6pt]
& \boxed{R_A = \frac{G}{1000}} \\[4pt]
& \text{Hence, the resistance of the ammeter is } \boxed{\tfrac{G}{1000}\ \Omega.}
\end{aligned}

$$\begin{gathered} I_g = \frac{I{R}_s}{R_s+R_g} \\ =\frac{0.1\times 1}{100+0.1} \\ {I}_g = 1 mA \end{gathered}$$

Galvanometer resistance,$R_g=50\Omega$, Initial series resistance, $R_{s1}=1k\Omega =1000\Omega$, Initial range, $V_1=2V$

The potential energy of a magnetic dipole in a magnetic field is:

$$U = -\vec{M} \cdot \vec{B} = -MB \cos\theta$$

where $\theta$ is the angle between $\vec{M}$ and $\vec{B}$.

Initially: $\theta_1 = 0°$

$$U_1 = -MB \cos 0° = -MB$$

Finally: $\theta_2 = 45°$

$$U_2 = -MB \cos 45° = -MB \frac{1}{\sqrt{2}}$$​

The work done by the external agent in rotating the dipole slowly (quasistatically) from $\theta_1$ to $\theta_2$ is equal to the increase in potential energy:

$$W = U_2 - U_1$$=​ $$W = \left[-MB \frac{1}{\sqrt{2}}\right] - (-MB)$$​​]

$$\boxed{W = MB \left(1 - \frac{1}{\sqrt{2}}\right)}$$​

$$\begin{gathered} r=\frac{mv}{eB}\&mvr=\frac{nh}{2\pi }\Rightarrow \left(eBr\right)r=\frac{nh}{2\pi } \\ \Rightarrow r=\sqrt{\frac{nh}{2\pi eB}} \\ \text{For the first excited state, }n=2\text{ : } \\ r=\sqrt{\frac{2h}{2\pi eB}}=\sqrt{\frac{h}{\pi eB}} \end{gathered}$$

\begin{aligned}
& \text{Given: } L = 10\ \text{cm} = 0.1\ \text{m}, \quad I = 1.0\ \text{A} \\[4pt]
& \text{Direction of current: along } +\hat{j} \\[2pt]
& \text{Magnetic field: } \vec{B} = (5\ \text{mT})\hat{j} - (8\ \text{mT})\hat{k}
= (5\times10^{-3})\hat{j} - (8\times10^{-3})\hat{k} \\[4pt]
& \text{Force on wire: } \vec{F} = I(\vec{L} \times \vec{B}) \\[4pt]
& \vec{L} = (0.1)\hat{j} \\[4pt]
& \Rightarrow \vec{F} = 1 \times \left[(0.1)\hat{j} \times \left(5\times10^{-3}\hat{j} - 8\times10^{-3}\hat{k}\right)\right] \\[4pt]
& \hat{j} \times \hat{j} = 0,\quad \hat{j} \times \hat{k} = \hat{i} \\[4pt]
& \Rightarrow \vec{F} = 0.1 \left[ -8\times10^{-3} (\hat{j} \times \hat{k}) \right]
= 0.1 \left[ -8\times10^{-3} \hat{i} \right] \\[4pt]
& \Rightarrow \vec{F} = -8\times10^{-4}\ \hat{i}\ \mathrm{N} \\[4pt]
& \text{Hence, the force on the wire is directed along } -x\text{-axis.} \\[6pt]
& \boxed{\vec{F} = -0.8\times10^{-3}\ \hat{i}\ \mathrm{N}}
\end{aligned}

$$\begin{gathered} r=\frac{mv}{eB}\&mvr=\frac{nh}{2\pi }\Rightarrow \left(eBr\right)r=\frac{nh}{2\pi } \\ \Rightarrow r=\sqrt{\frac{nh}{2\pi eB}} \\ \text{For the first excited state, }n=2\text{ : } \\ r=\sqrt{\frac{2h}{2\pi eB}}=\sqrt{\frac{h}{\pi eB}} \end{gathered}$$

\[
\begin{aligned}
&\text{Given:} \\
&I = 0.5~\mathrm{A}, \quad \vec{L} = 1~\mathrm{cm} = 0.01~\hat{i}~\mathrm{m}, \\
&\vec{B} = (0.4~\mathrm{mT})~\hat{j} + (0.6~\mathrm{mT})~\hat{k} = (0.4 \times 10^{-3})~\hat{j} + (0.6 \times 10^{-3})~\hat{k} \\[6pt]
&\text{Magnetic force: } \vec{F} = I(\vec{L} \times \vec{B}) \\[6pt]
&\vec{L} \times \vec{B} = (0.01\hat{i}) \times \left[(0.4\times10^{-3})\hat{j} + (0.6\times10^{-3})\hat{k}\right] \\[6pt]
&= 0.01\left[(0.4\times10^{-3})(\hat{i}\times\hat{j}) + (0.6\times10^{-3})(\hat{i}\times\hat{k})\right] \\[6pt]
&\text{Since } \hat{i}\times\hat{j} = \hat{k} \text{ and } \hat{i}\times\hat{k} = -\hat{j}, \\[4pt]
&\vec{L}\times\vec{B} = 0.01\left[(0.4\times10^{-3})\hat{k} - (0.6\times10^{-3})\hat{j}\right] \\[6pt]
&= (4\times10^{-6})\hat{k} - (6\times10^{-6})\hat{j} \\[6pt]
&\therefore \vec{F} = I(\vec{L}\times\vec{B}) = 0.5\left[(4\times10^{-6})\hat{k} - (6\times10^{-6})\hat{j}\right] \\[6pt]
&\boxed{\vec{F} = (-3\hat{j} + 2\hat{k})~\mu\mathrm{N}}
\end{aligned}
\]

$$\begin{gathered} I_g = \frac{I{R}_s}{R_s+R_g} \\ =\frac{0.1\times 1}{100+0.1} \\ {I}_g = 1 mA \end{gathered}$$

The potential energy of a magnetic dipole in a magnetic field is:

$$U = -\vec{M} \cdot \vec{B} = -MB \cos\theta$$

where $\theta$ is the angle between $\vec{M}$ and $\vec{B}$.

Initially: $\theta_1 = 0°$

$$U_1 = -MB \cos 0° = -MB$$

Finally: $\theta_2 = 45°$

$$U_2 = -MB \cos 45° = -MB \frac{1}{\sqrt{2}}$$​

The work done by the external agent in rotating the dipole slowly (quasistatically) from $\theta_1$ to $\theta_2$ is equal to the increase in potential energy:

$$W = U_2 - U_1$$=​ $$W = \left[-MB \frac{1}{\sqrt{2}}\right] - (-MB)$$​​]

$$\boxed{W = MB \left(1 - \frac{1}{\sqrt{2}}\right)}$$​

Radius of circular path is given by

$$\begin{gathered} \mathrm{r} = \frac{\mathrm{mv}}{\mathrm{qB}} \\ \mathrm{But}, \mathrm{p} =\mathrm{mv}=\sqrt{2\mathrm{mqV}} \\ \mathrm{So}, \\ \mathrm{r} = \frac{\sqrt{2\mathrm{mqV}}}{\mathrm{qB}} \\ \mathrm{r} \propto \sqrt{\mathrm{m}} \\ \mathrm{So}, \\ \frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}={\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^2 \end{gathered}$$

\begin{aligned}
& \text{Given: Galvanometer resistance } = G\ \Omega, \quad \text{Full-scale current } = I\ \mathrm{A}, \\[3pt]
& \text{Current through galvanometer } = 0.1\% \text{ of } I = \frac{0.1}{100}I = \frac{I}{1000}. \\[6pt]
& \text{Let shunt resistance } = S. \text{ Then the same potential difference acts across } G \text{ and } S. \\[3pt]
& \text{So, } I_g G = I_s S, \quad \text{where } I_s = I - I_g = I - \frac{I}{1000} = \frac{999I}{1000}. \\[4pt]
& \Rightarrow S = \frac{I_g G}{I_s} = \frac{\frac{I}{1000} G}{\frac{999I}{1000}} = \frac{G}{999}. \\[6pt]
& \text{The equivalent resistance of the ammeter is } R_A = G \parallel S \\[4pt]
& R_A = \frac{G \times S}{G + S} = \frac{G \times \frac{G}{999}}{G + \frac{G}{999}} = \frac{G^2 / 999}{G(1 + 1/999)} \\[4pt]
& R_A = \frac{G / 999}{1 + 1/999} = \frac{G / 999}{1000/999} = \frac{G}{1000}. \\[6pt]
& \boxed{R_A = \frac{G}{1000}} \\[4pt]
& \text{Hence, the resistance of the ammeter is } \boxed{\tfrac{G}{1000}\ \Omega.}
\end{aligned}

$$\begin{gathered} {\mathrm{I}}_{\mathrm{g}} = 0.1 \% \mathrm{of} \mathrm{I} = 0.001 \mathrm{I} \\ \mathrm{Current} \mathrm{through} \mathrm{shunt}, {\mathrm{I}}_{\mathrm{s}} = \mathrm{I} - {\mathrm{I}}_{\mathrm{g}} = \mathrm{I} - 0.001 \mathrm{I} = 0.999 \mathrm{I} \\ \mathrm{Let} \mathrm{S} \mathrm{is} \mathrm{the} \mathrm{shunt} \mathrm{resistance} \\ \mathrm{S} = \frac{{\mathrm{I}}_{\mathrm{g}} \mathrm{G}}{{\mathrm{I}}_{\mathrm{s}}}=\frac{0.001 \mathrm{I} \mathrm{G}}{0.999 \mathrm{I} }=\frac{\mathrm{G}}{999} \\ \mathrm{Resistance} \mathrm{of} \mathrm{ammeter} \mathrm{is} \mathrm{given} \mathrm{by} \\ \frac{1}{\mathrm{R}}=\frac{1}{\mathrm{G}}+\frac{1}{\mathrm{S}} \\ \frac{1}{\mathrm{R}}=\frac{1}{\mathrm{G}}+\frac{999}{\mathrm{G}}=\frac{1000}{\mathrm{G}} \\ \mathrm{R} = \frac{\mathrm{G}}{1000} \end{gathered}$$

$B = \frac{{\mu }_{0}i}{2\pi r}$

from Right hand thumb rule, direction of B is North.

Identify the direction of the current in the loop. Since the loop is in the xy-plane, the current flows in a circular path around the origin.

Determine the direction of the magnetic field. It is given that the magnetic field points along the +z-axis.

Apply the right-hand rule to find the direction of the force on each segment of the loop. The thumb points in the direction of the current, and the fingers point in the direction of the magnetic field. The force is perpendicular to both.

Since the magnetic field is uniform and the loop is symmetric, the forces on opposite segments of the loop will cancel each other out. Therefore, the net force on the loop is zero, and the loop will not move along any axis.

The loop will not move along any axis.

The potential energy of a magnetic dipole in a magnetic field is:

$$U = -\vec{M} \cdot \vec{B} = -MB \cos\theta$$

where $\theta$ is the angle between $\vec{M}$ and $\vec{B}$.

Initially: $\theta_1 = 0°$

$$U_1 = -MB \cos 0° = -MB$$

Finally: $\theta_2 = 45°$

$$U_2 = -MB \cos 45° = -MB \frac{1}{\sqrt{2}}$$​

The work done by the external agent in rotating the dipole slowly (quasistatically) from $\theta_1$ to $\theta_2$ is equal to the increase in potential energy:

$$W = U_2 - U_1$$=​ $$W = \left[-MB \frac{1}{\sqrt{2}}\right] - (-MB)$$​​]

$$\boxed{W = MB \left(1 - \frac{1}{\sqrt{2}}\right)}$$​

$$\begin{gathered} r=\frac{mv}{eB}\&mvr=\frac{nh}{2\pi }\Rightarrow \left(eBr\right)r=\frac{nh}{2\pi } \\ \Rightarrow r=\sqrt{\frac{nh}{2\pi eB}} \\ \text{For the first excited state, }n=2\text{ : } \\ r=\sqrt{\frac{2h}{2\pi eB}}=\sqrt{\frac{h}{\pi eB}} \end{gathered}$$

Galvanometer resistance,$R_g=50\Omega$, Initial series resistance, $R_{s1}=1k\Omega =1000\Omega$, Initial range, $V_1=2V$

$$\begin{gathered} r=\frac{mv}{eB}\&mvr=\frac{nh}{2\pi }\Rightarrow \left(eBr\right)r=\frac{nh}{2\pi } \\ \Rightarrow r=\sqrt{\frac{nh}{2\pi eB}} \\ \text{For the first excited state, }n=2\text{ : } \\ r=\sqrt{\frac{2h}{2\pi eB}}=\sqrt{\frac{h}{\pi eB}} \end{gathered}$$

Magnetic Force on a Current-Carrying Wire

The magnetic force (\(\vec{F}\)) on a straight wire of length \(L\) carrying a current \(I\) in a magnetic field \(\vec{B}\) is given by the formula:
\(\vec{F} = I (\vec{L} \times \vec{B})\)

Current: \(I = 2.0 \text{ A}\)
Length: \(L = 1.0 \text{ m}\)
Magnetic Field: \(\vec{B} = (3\hat{i} + 2\hat{j}) \text{ T}\)
Length Vector (\(\vec{L}\) is along \(+x\) direction): \(\vec{L} = L\hat{i} = (1.0 \text{ m})\hat{i}\)
Calculation

Step 1: Calculate the cross product \(\vec{L} \times \vec{B}\)
\(\vec{L} \times \vec{B} = (1.0\hat{i}) \times (3\hat{i} + 2\hat{j})\)
\(\vec{L} \times \vec{B} = 3(\hat{i} \times \hat{i}) + 2(\hat{i} \times \hat{j})\)
Since \(\hat{i} \times \hat{i} = 0\) and \(\hat{i} \times \hat{j} = \hat{k}\):
\(\vec{L} \times \vec{B} = 0 + 2\hat{k}\)
\(\vec{L} \times \vec{B} = 2\hat{k} \text{ T}\cdot\text{m}\)

Step 2: Calculate the force \(\vec{F}\)
\(\vec{F} = I (\vec{L} \times \vec{B})\)
\(\vec{F} = (2.0 \text{ A}) (2\hat{k} \text{ T}\cdot\text{m})\)
\(\vec{F} = 4.0\hat{k} \text{ N}\)

Final Answer
The magnitude of the force is \(4.0 \text{ N}\) and the direction is along the \(\textbf{+z-axis}\).

This corresponds to Option C.

Given: A loop carrying current I clockwise when viewed from +x-axis.
Hence, the magnetic moment $\mu$ is along -x.
Magnetic field B is along +z.
Each small current element Idl on the loop experiences

$F =Idl\times B$
Since the loop lies in the zy-plane, B (along z) is perpendicular to some elements and parallel to others.
For an element where current flows along +y,$F =Idl\times B$ is along +x.
For the opposite element (current along -y ), F is along -x.
These forces act outward on opposite sides - tending to pull the loop outward.
Hence, the loop experiences an outward (radial) magnetic pressure.

$$\begin{gathered} \overrightarrow{B}=\left(5\mathrm{mT}\right)\hat{j}-\left(8\mathrm{mT}\right)\hat{k} \\ \mathrm{I} = 1 \mathrm{A}, \overrightarrow{\mathrm{L}} = 0.1 \hat{\mathrm{j}} \mathrm{m} \\ \overrightarrow{\mathrm{F}} = \mathrm{I}\left(\overrightarrow{\mathrm{L}} \times \overrightarrow{\mathrm{B}}\right) \\ \overrightarrow{\mathrm{F}} = 1\left(0.1 \hat{\mathrm{j}} \times \left(5\hat{j}-8\hat{k} \right){10}^{-3}\right) \\ \overrightarrow{\mathrm{F}} = - 0.8 \hat{\mathrm{i}} \mathrm{mN} \end{gathered}$$

Magnetic Force on a Current-Carrying Wire

The magnetic force (\(\vec{F}\)) on a straight wire of length \(L\) carrying a current \(I\) in a magnetic field \(\vec{B}\) is given by the formula:
\(\vec{F} = I (\vec{L} \times \vec{B})\)

Current: \(I = 2.0 \text{ A}\)
Length: \(L = 1.0 \text{ m}\)
Magnetic Field: \(\vec{B} = (3\hat{i} + 2\hat{j}) \text{ T}\)
Length Vector (\(\vec{L}\) is along \(+x\) direction): \(\vec{L} = L\hat{i} = (1.0 \text{ m})\hat{i}\)
Calculation

Step 1: Calculate the cross product \(\vec{L} \times \vec{B}\)
\(\vec{L} \times \vec{B} = (1.0\hat{i}) \times (3\hat{i} + 2\hat{j})\)
\(\vec{L} \times \vec{B} = 3(\hat{i} \times \hat{i}) + 2(\hat{i} \times \hat{j})\)
Since \(\hat{i} \times \hat{i} = 0\) and \(\hat{i} \times \hat{j} = \hat{k}\):
\(\vec{L} \times \vec{B} = 0 + 2\hat{k}\)
\(\vec{L} \times \vec{B} = 2\hat{k} \text{ T}\cdot\text{m}\)

Step 2: Calculate the force \(\vec{F}\)
\(\vec{F} = I (\vec{L} \times \vec{B})\)
\(\vec{F} = (2.0 \text{ A}) (2\hat{k} \text{ T}\cdot\text{m})\)
\(\vec{F} = 4.0\hat{k} \text{ N}\)

Final Answer
The magnitude of the force is \(4.0 \text{ N}\) and the direction is along the \(\textbf{+z-axis}\).

This corresponds to Option C.

$$\begin{gathered} I_g = \frac{I{R}_s}{R_s+R_g} \\ =\frac{0.1\times 1}{100+0.1} \\ {I}_g = 1 mA \end{gathered}$$

दिया गया है: +x-अक्ष से देखने पर दक्षिणावर्त दिशा में धारा I ले जाने वाला एक लूप।
अतः, चुंबकीय आघूर्ण ऋणात्मक x-अक्ष (-x) के अनुदिश है।
चुंबकीय क्षेत्र , धनात्मक z-अक्ष (+z) के अनुदिश है।
लूप पर प्रत्येक छोटा धारा अवयव Idl निम्नलिखित बल का अनुभव करता है:
$\overrightarrow{F}=Id\overrightarrow{l}\times \overrightarrow{B}$

चूँकि लूप zy-तल में स्थित है, इसलिए $\overrightarrow{B}$ (z के अनुदिश) कुछ अवयवों के लंबवत है और अन्य के समांतर है।
एक अवयव के लिए जहाँ धारा +y के अनुदिश प्रवाहित होती है, बल $\overrightarrow{F}=Id\overrightarrow{l}\times \overrightarrow{B}$, के अनुदिश कार्य करता है।
इसके विपरीत अवयव (धारा -y के अनुदिश) के लिए, बल $\overrightarrow{F}$ , -x के अनुदिश कार्य करता है।
ये बल विपरीत पक्षों पर बाहर की ओर कार्य करते हैं - जिससे लूप बाहर की ओर खिंचता है।
अतः, लूप एक बाहरी (त्रिज्यीय) चुंबकीय दाब का अनुभव करता है, जिससे इसकी प्रवृत्ति फैलने की होगी।

Galvanometer resistance,$R_g=50\Omega$, Initial series resistance, $R_{s1}=1k\Omega =1000\Omega$, Initial range, $V_1=2V$

Radius of circular path is given by

$$\begin{gathered} \mathrm{r} = \frac{\mathrm{mv}}{\mathrm{qB}} \\ \mathrm{But}, \mathrm{p} =\mathrm{mv}=\sqrt{2\mathrm{mqV}} \\ \mathrm{So}, \\ \mathrm{r} = \frac{\sqrt{2\mathrm{mqV}}}{\mathrm{qB}} \\ \mathrm{r} \propto \sqrt{\mathrm{m}} \\ \mathrm{So}, \\ \frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}={\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^2 \end{gathered}$$

\[
\begin{aligned}
&\text{Given:} \\
&I = 0.5~\mathrm{A}, \quad \vec{L} = 1~\mathrm{cm} = 0.01~\hat{i}~\mathrm{m}, \\
&\vec{B} = (0.4~\mathrm{mT})~\hat{j} + (0.6~\mathrm{mT})~\hat{k} = (0.4 \times 10^{-3})~\hat{j} + (0.6 \times 10^{-3})~\hat{k} \\[6pt]
&\text{Magnetic force: } \vec{F} = I(\vec{L} \times \vec{B}) \\[6pt]
&\vec{L} \times \vec{B} = (0.01\hat{i}) \times \left[(0.4\times10^{-3})\hat{j} + (0.6\times10^{-3})\hat{k}\right] \\[6pt]
&= 0.01\left[(0.4\times10^{-3})(\hat{i}\times\hat{j}) + (0.6\times10^{-3})(\hat{i}\times\hat{k})\right] \\[6pt]
&\text{Since } \hat{i}\times\hat{j} = \hat{k} \text{ and } \hat{i}\times\hat{k} = -\hat{j}, \\[4pt]
&\vec{L}\times\vec{B} = 0.01\left[(0.4\times10^{-3})\hat{k} - (0.6\times10^{-3})\hat{j}\right] \\[6pt]
&= (4\times10^{-6})\hat{k} - (6\times10^{-6})\hat{j} \\[6pt]
&\therefore \vec{F} = I(\vec{L}\times\vec{B}) = 0.5\left[(4\times10^{-6})\hat{k} - (6\times10^{-6})\hat{j}\right] \\[6pt]
&\boxed{\vec{F} = (-3\hat{j} + 2\hat{k})~\mu\mathrm{N}}
\end{aligned}
\]

Given: A loop carrying current I clockwise when viewed from +x-axis.
Hence, the magnetic moment $\mu$ is along -x.
Magnetic field B is along +z.
Each small current element Idl on the loop experiences

$F =Idl\times B$
Since the loop lies in the zy-plane, B (along z) is perpendicular to some elements and parallel to others.
For an element where current flows along +y,$F =Idl\times B$ is along +x.
For the opposite element (current along -y ), F is along -x.
These forces act outward on opposite sides - tending to pull the loop outward.
Hence, the loop experiences an outward (radial) magnetic pressure.

$$\begin{gathered} \overrightarrow{B}=\left(5\mathrm{mT}\right)\hat{j}-\left(8\mathrm{mT}\right)\hat{k} \\ \mathrm{I} = 1 \mathrm{A}, \overrightarrow{\mathrm{L}} = 0.1 \hat{\mathrm{j}} \mathrm{m} \\ \overrightarrow{\mathrm{F}} = \mathrm{I}\left(\overrightarrow{\mathrm{L}} \times \overrightarrow{\mathrm{B}}\right) \\ \overrightarrow{\mathrm{F}} = 1\left(0.1 \hat{\mathrm{j}} \times \left(5\hat{j}-8\hat{k} \right){10}^{-3}\right) \\ \overrightarrow{\mathrm{F}} = - 0.8 \hat{\mathrm{i}} \mathrm{mN} \end{gathered}$$

The potential energy of a magnetic dipole in a magnetic field is:

$$U = -\vec{M} \cdot \vec{B} = -MB \cos\theta$$

where $\theta$ is the angle between $\vec{M}$ and $\vec{B}$.

Initially: $\theta_1 = 0°$

$$U_1 = -MB \cos 0° = -MB$$

Finally: $\theta_2 = 45°$

$$U_2 = -MB \cos 45° = -MB \frac{1}{\sqrt{2}}$$​

The work done by the external agent in rotating the dipole slowly (quasistatically) from $\theta_1$ to $\theta_2$ is equal to the increase in potential energy:

$$W = U_2 - U_1$$=​ $$W = \left[-MB \frac{1}{\sqrt{2}}\right] - (-MB)$$​​]

$$\boxed{W = MB \left(1 - \frac{1}{\sqrt{2}}\right)}$$​

\begin{aligned}
& \text{Given: Galvanometer resistance } = G\ \Omega, \quad \text{Full-scale current } = I\ \mathrm{A}, \\[3pt]
& \text{Current through galvanometer } = 0.1\% \text{ of } I = \frac{0.1}{100}I = \frac{I}{1000}. \\[6pt]
& \text{Let shunt resistance } = S. \text{ Then the same potential difference acts across } G \text{ and } S. \\[3pt]
& \text{So, } I_g G = I_s S, \quad \text{where } I_s = I - I_g = I - \frac{I}{1000} = \frac{999I}{1000}. \\[4pt]
& \Rightarrow S = \frac{I_g G}{I_s} = \frac{\frac{I}{1000} G}{\frac{999I}{1000}} = \frac{G}{999}. \\[6pt]
& \text{The equivalent resistance of the ammeter is } R_A = G \parallel S \\[4pt]
& R_A = \frac{G \times S}{G + S} = \frac{G \times \frac{G}{999}}{G + \frac{G}{999}} = \frac{G^2 / 999}{G(1 + 1/999)} \\[4pt]
& R_A = \frac{G / 999}{1 + 1/999} = \frac{G / 999}{1000/999} = \frac{G}{1000}. \\[6pt]
& \boxed{R_A = \frac{G}{1000}} \\[4pt]
& \text{Hence, the resistance of the ammeter is } \boxed{\tfrac{G}{1000}\ \Omega.}
\end{aligned}

Both (A) and (R) are true and (R) Is the correct explanation Of (A)

$\overline{\mathrm{F}}=\mathrm{q}(\overline{\mathrm{v}}\times \overline{\mathrm{B}})$

$\overline{\mathrm{F}}=0$

$\overline{\mathrm{V}}\Vert \overline{\mathrm{B}}$

$\theta =0\text{ or }180$

Identify the direction of the current in the loop. Since the loop is in the xy-plane, the current flows in a circular path around the origin.

Determine the direction of the magnetic field. It is given that the magnetic field points along the +z-axis.

Apply the right-hand rule to find the direction of the force on each segment of the loop. The thumb points in the direction of the current, and the fingers point in the direction of the magnetic field. The force is perpendicular to both.

Since the magnetic field is uniform and the loop is symmetric, the forces on opposite segments of the loop will cancel each other out. Therefore, the net force on the loop is zero, and the loop will not move along any axis.

The loop will not move along any axis.

Magnetic Force on a Current-Carrying Wire

The magnetic force (\(\vec{F}\)) on a straight wire of length \(L\) carrying a current \(I\) in a magnetic field \(\vec{B}\) is given by the formula:
\(\vec{F} = I (\vec{L} \times \vec{B})\)

Current: \(I = 2.0 \text{ A}\)
Length: \(L = 1.0 \text{ m}\)
Magnetic Field: \(\vec{B} = (3\hat{i} + 2\hat{j}) \text{ T}\)
Length Vector (\(\vec{L}\) is along \(+x\) direction): \(\vec{L} = L\hat{i} = (1.0 \text{ m})\hat{i}\)
Calculation

Step 1: Calculate the cross product \(\vec{L} \times \vec{B}\)
\(\vec{L} \times \vec{B} = (1.0\hat{i}) \times (3\hat{i} + 2\hat{j})\)
\(\vec{L} \times \vec{B} = 3(\hat{i} \times \hat{i}) + 2(\hat{i} \times \hat{j})\)
Since \(\hat{i} \times \hat{i} = 0\) and \(\hat{i} \times \hat{j} = \hat{k}\):
\(\vec{L} \times \vec{B} = 0 + 2\hat{k}\)
\(\vec{L} \times \vec{B} = 2\hat{k} \text{ T}\cdot\text{m}\)

Step 2: Calculate the force \(\vec{F}\)
\(\vec{F} = I (\vec{L} \times \vec{B})\)
\(\vec{F} = (2.0 \text{ A}) (2\hat{k} \text{ T}\cdot\text{m})\)
\(\vec{F} = 4.0\hat{k} \text{ N}\)

Final Answer
The magnitude of the force is \(4.0 \text{ N}\) and the direction is along the \(\textbf{+z-axis}\).

This corresponds to Option C.

$$\begin{gathered} \overrightarrow{B}=\left(5\mathrm{mT}\right)\hat{j}-\left(8\mathrm{mT}\right)\hat{k} \\ \mathrm{I} = 1 \mathrm{A}, \overrightarrow{\mathrm{L}} = 0.1 \hat{\mathrm{j}} \mathrm{m} \\ \overrightarrow{\mathrm{F}} = \mathrm{I}\left(\overrightarrow{\mathrm{L}} \times \overrightarrow{\mathrm{B}}\right) \\ \overrightarrow{\mathrm{F}} = 1\left(0.1 \hat{\mathrm{j}} \times \left(5\hat{j}-8\hat{k} \right){10}^{-3}\right) \\ \overrightarrow{\mathrm{F}} = - 0.8 \hat{\mathrm{i}} \mathrm{mN} \end{gathered}$$

\begin{aligned}
& \text{Given: Galvanometer resistance } = G\ \Omega, \quad \text{Full-scale current } = I\ \mathrm{A}, \\[3pt]
& \text{Current through galvanometer } = 0.1\% \text{ of } I = \frac{0.1}{100}I = \frac{I}{1000}. \\[6pt]
& \text{Let shunt resistance } = S. \text{ Then the same potential difference acts across } G \text{ and } S. \\[3pt]
& \text{So, } I_g G = I_s S, \quad \text{where } I_s = I - I_g = I - \frac{I}{1000} = \frac{999I}{1000}. \\[4pt]
& \Rightarrow S = \frac{I_g G}{I_s} = \frac{\frac{I}{1000} G}{\frac{999I}{1000}} = \frac{G}{999}. \\[6pt]
& \text{The equivalent resistance of the ammeter is } R_A = G \parallel S \\[4pt]
& R_A = \frac{G \times S}{G + S} = \frac{G \times \frac{G}{999}}{G + \frac{G}{999}} = \frac{G^2 / 999}{G(1 + 1/999)} \\[4pt]
& R_A = \frac{G / 999}{1 + 1/999} = \frac{G / 999}{1000/999} = \frac{G}{1000}. \\[6pt]
& \boxed{R_A = \frac{G}{1000}} \\[4pt]
& \text{Hence, the resistance of the ammeter is } \boxed{\tfrac{G}{1000}\ \Omega.}
\end{aligned}

1 : 4

Reason: Magnetic moment $m = IA = I\pi r^2$.
With the same current, $m \propto r^2$.
So $m_1:m_2 = r_1^2:r_2^2 = 1^2:2^2 = 1:4$.

\begin{aligned}
& \text{Given: } L = 10\ \text{cm} = 0.1\ \text{m}, \quad I = 1.0\ \text{A} \\[4pt]
& \text{Direction of current: along } +\hat{j} \\[2pt]
& \text{Magnetic field: } \vec{B} = (5\ \text{mT})\hat{j} - (8\ \text{mT})\hat{k}
= (5\times10^{-3})\hat{j} - (8\times10^{-3})\hat{k} \\[4pt]
& \text{Force on wire: } \vec{F} = I(\vec{L} \times \vec{B}) \\[4pt]
& \vec{L} = (0.1)\hat{j} \\[4pt]
& \Rightarrow \vec{F} = 1 \times \left[(0.1)\hat{j} \times \left(5\times10^{-3}\hat{j} - 8\times10^{-3}\hat{k}\right)\right] \\[4pt]
& \hat{j} \times \hat{j} = 0,\quad \hat{j} \times \hat{k} = \hat{i} \\[4pt]
& \Rightarrow \vec{F} = 0.1 \left[ -8\times10^{-3} (\hat{j} \times \hat{k}) \right]
= 0.1 \left[ -8\times10^{-3} \hat{i} \right] \\[4pt]
& \Rightarrow \vec{F} = -8\times10^{-4}\ \hat{i}\ \mathrm{N} \\[4pt]
& \text{Hence, the force on the wire is directed along } -x\text{-axis.} \\[6pt]
& \boxed{\vec{F} = -0.8\times10^{-3}\ \hat{i}\ \mathrm{N}}
\end{aligned}

The final speed of the electron is v, the same as its initial speed. This is because the magnetic force that causes the semicircular path is always perpendicular to the electron's velocity, meaning it does no work on the electron and does not change its kinetic energy or speed.

Identify the direction of the current in the loop. Since the loop is in the xy-plane, the current flows in a circular path around the origin.

Determine the direction of the magnetic field. It is given that the magnetic field points along the +z-axis.

Apply the right-hand rule to find the direction of the force on each segment of the loop. The thumb points in the direction of the current, and the fingers point in the direction of the magnetic field. The force is perpendicular to both.

Since the magnetic field is uniform and the loop is symmetric, the forces on opposite segments of the loop will cancel each other out. Therefore, the net force on the loop is zero, and the loop will not move along any axis.

The loop will not move along any axis.

To find the magnetic field at the centroid of an equilateral triangle frame of side l carrying a current i, we use the Biot-Savart law.

For a current-carrying finite wire, the magnetic field at a perpendicular distance d from the center of the wire is:

$$B = \frac{\mu_0 i}{4 \pi d} \left[ \sin \theta_1 + \sin \theta_2 \right]$$

For an equilateral triangle, the perpendicular distance from the centroid to any side is:

$d=\frac{l}{2\sqrt{3}}$​

Each side of the equilateral triangle subtends an angle θ = 60° at the centroid.

So,

$$\sin 60^\circ = \frac{\sqrt{3}}{2}$$​

Thus, the field due to one side:

$B_{\text{one side}}=\frac{{\mu }_{0}i}{4\pi \left(l\mathrm{/}2\sqrt{3}\right)}(2\times \frac{\sqrt{3}}{2})$

​​$B_{\text{one side}}=\frac{{\mu }_{0}i}{2\pi }\times \frac{\sqrt{3}}{l}\times \sqrt{3}$

​ $B_{\text{one side}}=\frac{{\mu }_{0}i}{2\pi l}\times 3$

$B_{\text{one side}}=\frac{3{\mu }_{0}i}{2\pi l}$​

Since all three sides contribute equally to the field at the centroid and add up vectorially:

$$B_{\text{total}} = 3 \times B_{\text{one side}}$$

​ $B_{\text{total}}=3\times \frac{3{\mu }_{0}i}{2\pi l}$

$B_{\text{total}}=\frac{9{\mu }_{0}i}{2\pi l}$​

From the given options:

• The correct answer is Option A: $\frac{9{\mu }_{0}i}{2\pi l}$​

Option C: $\frac{9{\mu }_{0}i}{2\pi l}$​

Radius of circular path is given by

$$\begin{gathered} \mathrm{r} = \frac{\mathrm{mv}}{\mathrm{qB}} \\ \mathrm{But}, \mathrm{p} =\mathrm{mv}=\sqrt{2\mathrm{mqV}} \\ \mathrm{So}, \\ \mathrm{r} = \frac{\sqrt{2\mathrm{mqV}}}{\mathrm{qB}} \\ \mathrm{r} \propto \sqrt{\mathrm{m}} \\ \mathrm{So}, \\ \frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}={\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^2 \end{gathered}$$

$$\begin{gathered} {\mathrm{I}}_{\mathrm{g}} = 0.1 \% \mathrm{of} \mathrm{I} = 0.001 \mathrm{I} \\ \mathrm{Current} \mathrm{through} \mathrm{shunt}, {\mathrm{I}}_{\mathrm{s}} = \mathrm{I} - {\mathrm{I}}_{\mathrm{g}} = \mathrm{I} - 0.001 \mathrm{I} = 0.999 \mathrm{I} \\ \mathrm{Let} \mathrm{S} \mathrm{is} \mathrm{the} \mathrm{shunt} \mathrm{resistance} \\ \mathrm{S} = \frac{{\mathrm{I}}_{\mathrm{g}} \mathrm{G}}{{\mathrm{I}}_{\mathrm{s}}}=\frac{0.001 \mathrm{I} \mathrm{G}}{0.999 \mathrm{I} }=\frac{\mathrm{G}}{999} \\ \mathrm{Resistance} \mathrm{of} \mathrm{ammeter} \mathrm{is} \mathrm{given} \mathrm{by} \\ \frac{1}{\mathrm{R}}=\frac{1}{\mathrm{G}}+\frac{1}{\mathrm{S}} \\ \frac{1}{\mathrm{R}}=\frac{1}{\mathrm{G}}+\frac{999}{\mathrm{G}}=\frac{1000}{\mathrm{G}} \\ \mathrm{R} = \frac{\mathrm{G}}{1000} \end{gathered}$$

$$\begin{gathered} {\mathrm{I}}_{\mathrm{g}} = 0.1 \% \mathrm{of} \mathrm{I} = 0.001 \mathrm{I} \\ \mathrm{Current} \mathrm{through} \mathrm{shunt}, {\mathrm{I}}_{\mathrm{s}} = \mathrm{I} - {\mathrm{I}}_{\mathrm{g}} = \mathrm{I} - 0.001 \mathrm{I} = 0.999 \mathrm{I} \\ \mathrm{Let} \mathrm{S} \mathrm{is} \mathrm{the} \mathrm{shunt} \mathrm{resistance} \\ \mathrm{S} = \frac{{\mathrm{I}}_{\mathrm{g}} \mathrm{G}}{{\mathrm{I}}_{\mathrm{s}}}=\frac{0.001 \mathrm{I} \mathrm{G}}{0.999 \mathrm{I} }=\frac{\mathrm{G}}{999} \\ \mathrm{Resistance} \mathrm{of} \mathrm{ammeter} \mathrm{is} \mathrm{given} \mathrm{by} \\ \frac{1}{\mathrm{R}}=\frac{1}{\mathrm{G}}+\frac{1}{\mathrm{S}} \\ \frac{1}{\mathrm{R}}=\frac{1}{\mathrm{G}}+\frac{999}{\mathrm{G}}=\frac{1000}{\mathrm{G}} \\ \mathrm{R} = \frac{\mathrm{G}}{1000} \end{gathered}$$

The potential energy of a magnetic dipole in a magnetic field is:

$$U = -\vec{M} \cdot \vec{B} = -MB \cos\theta$$

where $\theta$ is the angle between $\vec{M}$ and $\vec{B}$.

Initially: $\theta_1 = 0°$

$$U_1 = -MB \cos 0° = -MB$$

Finally: $\theta_2 = 45°$

$$U_2 = -MB \cos 45° = -MB \frac{1}{\sqrt{2}}$$​

The work done by the external agent in rotating the dipole slowly (quasistatically) from $\theta_1$ to $\theta_2$ is equal to the increase in potential energy:

$$W = U_2 - U_1$$=​ $$W = \left[-MB \frac{1}{\sqrt{2}}\right] - (-MB)$$​​]

$$\boxed{W = MB \left(1 - \frac{1}{\sqrt{2}}\right)}$$​