NEET Test Series from KOTA - 10 Papers In MS WORD
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Moving Charges & Magnetism
153799
Two thin long parallel wires separated by a distance $b$ meter are carrying a current of $I$ ampere each. The magnitude of the force per unit length exerted by one wire on the other is [ $\mu_{0}=$ permeability constant]
C Given, Distance between two wire $=\mathrm{b}$ meter \(\begin{array}{ll} & \text { Current }=\text { IA } \\ \text { Let, } & \text { Length }=l\end{array}\) We know that, $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}$ $\mathrm{B}_{\mathrm{A}}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{b}}$ Force of wire due to field of A- $\mathrm{F}=\mathrm{IB}_{\mathrm{A}} \cdot l$ $\mathrm{~F}=\mathrm{I} \frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{b}} l$ $\frac{\mathrm{F}}{l}=\frac{\mu_{0} \mathrm{I}^{2}}{2 \pi \mathrm{b}} \mathrm{N} / \mathrm{m}$ Hence, magnitude of the force per unit length exerted by one wire on the other is $\frac{\mu_{0} I^{2}}{2 \pi b} \mathrm{~N} / \mathrm{m}$.
CG PET- 2004
Moving Charges & Magnetism
153800
A wire of length $\ell$ is bent to the form of a circular coil of some turns. A current i flows through the coil. The coil is placed in a uniform magnetic field $B$. The maximum torque on the coil can be
1 $\frac{\mathrm{iB} \ell^{2}}{4 \pi}$
2 $\frac{\mathrm{iB} \ell^{2}}{\pi}$
3 $\frac{\mathrm{iB} \ell^{2}}{2 \pi}$
4 $\frac{2 \mathrm{iB} \ell^{2}}{\pi}$
Explanation:
A We know that, $l=2 \pi \mathrm{RN}$ Where, $\mathrm{N}=\text { Number of turns }$ $\mathrm{R}=\text { Radius of the coil }$ $\mathrm{R}=\frac{l}{2 \pi \mathrm{N}}$ Magnetic moment $\mathrm{M}=\mathrm{NiA}=\mathrm{Ni}\left(\pi \mathrm{R}^{2}\right)$ $=\mathrm{Ni}\left(\frac{\pi l^{2}}{4 \pi^{2} \mathrm{~N}^{2}}\right)=\frac{\mathrm{i} l^{2}}{4 \pi \mathrm{N}}$ Maximum value of $\mathrm{M}$ can be, $\mathrm{M}_{\text {max }}=\frac{\mathrm{i} l^{2}}{4 \pi} \quad(\because \mathrm{N}=1)$ $\tau_{\max }=\mathrm{M}_{\max } \mathrm{B} \sin 90^{\circ}=\frac{\mathrm{iB} l^{2}}{4 \pi}$
BITSAT-2015
Moving Charges & Magnetism
153801
Two parallel wires carrying currents $I_{1}$ and $I_{2}$ in opposite directions and separated by a distance $d$ experience a
1 Repulsive force $\mu_{0} I_{1} I_{2} / 2 \pi d$
2 Attractive force $\mu_{0} I_{1} I_{2} / 2 \pi d$
3 Repulsive force $\mu_{0} \mathrm{I}_{1} \mathrm{I}_{2} / 2 \pi \mathrm{d}^{2}$
4 Attractive force $\mu_{0} I_{1} I_{2} / 2 \pi d^{2}$
Explanation:
A When two wires carrying currents $\mathrm{I}_{1} \& \mathrm{I}_{2}$ in opposite direction then they will repel each other. Magnetic field created by current in (1) wire. $\mathrm{B}_{1}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \pi \mathrm{d}}$ Now magnetic force on wire (2) due to wire (1) per unit length $\mathrm{F}_{21}=\mathrm{I}_{2} l \mathrm{~B}_{1}$ $\mathrm{~F}_{21}=\mathrm{I}_{2} l \times \frac{\mu_{0} \mathrm{I}_{1}}{2 \pi \mathrm{d}}$ $\frac{\mathrm{F}_{21}}{l}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1} \mathrm{I}_{2}}{2 \pi \mathrm{d}}$ $\mathrm{F}_{21}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1} \mathrm{I}_{2}}{2 \pi \mathrm{d}} \quad(\therefore l=1)$ Then, $\quad F_{21}=F_{12}=F=\frac{\mu_{0} I_{1} I_{2}}{2 \pi d}$ As the direction of currents are opposite, the force is repulsive force
UPSEE - 2011
Moving Charges & Magnetism
153802
The wire of length $l$ is bent into a circular loop of a single turn and is suspended in a magnetic field of induction $B$. When a current $I$ is passed through the loop, the maximum torque experienced by it is
A Let the radius of circular loop be $r$ then circumference of circular loop $l=2 \pi \mathrm{r} \quad \text { or } \quad \mathrm{r}=\frac{l}{2 \pi}$ When current $I$ is passed through the loop, the maximum torque acting on the loop is given by $\tau =\mathrm{I} \mathrm{B} \mathrm{A}$ $\tau =\mathrm{I} \mathrm{B} \times \pi \mathrm{r}^{2}$ $\tau =\mathrm{IB} \pi \times\left(\frac{l}{2 \pi}\right)^{2} \quad\left(\because \mathrm{r}=\frac{l}{2 \pi}\right)$ $\tau =\mathrm{IB} \pi \times \frac{l^{2}}{4 \pi^{2}}$ $\tau =\left(\frac{1}{4 \pi}\right) \mathrm{BI} l^{2}$
153799
Two thin long parallel wires separated by a distance $b$ meter are carrying a current of $I$ ampere each. The magnitude of the force per unit length exerted by one wire on the other is [ $\mu_{0}=$ permeability constant]
C Given, Distance between two wire $=\mathrm{b}$ meter \(\begin{array}{ll} & \text { Current }=\text { IA } \\ \text { Let, } & \text { Length }=l\end{array}\) We know that, $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}$ $\mathrm{B}_{\mathrm{A}}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{b}}$ Force of wire due to field of A- $\mathrm{F}=\mathrm{IB}_{\mathrm{A}} \cdot l$ $\mathrm{~F}=\mathrm{I} \frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{b}} l$ $\frac{\mathrm{F}}{l}=\frac{\mu_{0} \mathrm{I}^{2}}{2 \pi \mathrm{b}} \mathrm{N} / \mathrm{m}$ Hence, magnitude of the force per unit length exerted by one wire on the other is $\frac{\mu_{0} I^{2}}{2 \pi b} \mathrm{~N} / \mathrm{m}$.
CG PET- 2004
Moving Charges & Magnetism
153800
A wire of length $\ell$ is bent to the form of a circular coil of some turns. A current i flows through the coil. The coil is placed in a uniform magnetic field $B$. The maximum torque on the coil can be
1 $\frac{\mathrm{iB} \ell^{2}}{4 \pi}$
2 $\frac{\mathrm{iB} \ell^{2}}{\pi}$
3 $\frac{\mathrm{iB} \ell^{2}}{2 \pi}$
4 $\frac{2 \mathrm{iB} \ell^{2}}{\pi}$
Explanation:
A We know that, $l=2 \pi \mathrm{RN}$ Where, $\mathrm{N}=\text { Number of turns }$ $\mathrm{R}=\text { Radius of the coil }$ $\mathrm{R}=\frac{l}{2 \pi \mathrm{N}}$ Magnetic moment $\mathrm{M}=\mathrm{NiA}=\mathrm{Ni}\left(\pi \mathrm{R}^{2}\right)$ $=\mathrm{Ni}\left(\frac{\pi l^{2}}{4 \pi^{2} \mathrm{~N}^{2}}\right)=\frac{\mathrm{i} l^{2}}{4 \pi \mathrm{N}}$ Maximum value of $\mathrm{M}$ can be, $\mathrm{M}_{\text {max }}=\frac{\mathrm{i} l^{2}}{4 \pi} \quad(\because \mathrm{N}=1)$ $\tau_{\max }=\mathrm{M}_{\max } \mathrm{B} \sin 90^{\circ}=\frac{\mathrm{iB} l^{2}}{4 \pi}$
BITSAT-2015
Moving Charges & Magnetism
153801
Two parallel wires carrying currents $I_{1}$ and $I_{2}$ in opposite directions and separated by a distance $d$ experience a
1 Repulsive force $\mu_{0} I_{1} I_{2} / 2 \pi d$
2 Attractive force $\mu_{0} I_{1} I_{2} / 2 \pi d$
3 Repulsive force $\mu_{0} \mathrm{I}_{1} \mathrm{I}_{2} / 2 \pi \mathrm{d}^{2}$
4 Attractive force $\mu_{0} I_{1} I_{2} / 2 \pi d^{2}$
Explanation:
A When two wires carrying currents $\mathrm{I}_{1} \& \mathrm{I}_{2}$ in opposite direction then they will repel each other. Magnetic field created by current in (1) wire. $\mathrm{B}_{1}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \pi \mathrm{d}}$ Now magnetic force on wire (2) due to wire (1) per unit length $\mathrm{F}_{21}=\mathrm{I}_{2} l \mathrm{~B}_{1}$ $\mathrm{~F}_{21}=\mathrm{I}_{2} l \times \frac{\mu_{0} \mathrm{I}_{1}}{2 \pi \mathrm{d}}$ $\frac{\mathrm{F}_{21}}{l}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1} \mathrm{I}_{2}}{2 \pi \mathrm{d}}$ $\mathrm{F}_{21}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1} \mathrm{I}_{2}}{2 \pi \mathrm{d}} \quad(\therefore l=1)$ Then, $\quad F_{21}=F_{12}=F=\frac{\mu_{0} I_{1} I_{2}}{2 \pi d}$ As the direction of currents are opposite, the force is repulsive force
UPSEE - 2011
Moving Charges & Magnetism
153802
The wire of length $l$ is bent into a circular loop of a single turn and is suspended in a magnetic field of induction $B$. When a current $I$ is passed through the loop, the maximum torque experienced by it is
A Let the radius of circular loop be $r$ then circumference of circular loop $l=2 \pi \mathrm{r} \quad \text { or } \quad \mathrm{r}=\frac{l}{2 \pi}$ When current $I$ is passed through the loop, the maximum torque acting on the loop is given by $\tau =\mathrm{I} \mathrm{B} \mathrm{A}$ $\tau =\mathrm{I} \mathrm{B} \times \pi \mathrm{r}^{2}$ $\tau =\mathrm{IB} \pi \times\left(\frac{l}{2 \pi}\right)^{2} \quad\left(\because \mathrm{r}=\frac{l}{2 \pi}\right)$ $\tau =\mathrm{IB} \pi \times \frac{l^{2}}{4 \pi^{2}}$ $\tau =\left(\frac{1}{4 \pi}\right) \mathrm{BI} l^{2}$
153799
Two thin long parallel wires separated by a distance $b$ meter are carrying a current of $I$ ampere each. The magnitude of the force per unit length exerted by one wire on the other is [ $\mu_{0}=$ permeability constant]
C Given, Distance between two wire $=\mathrm{b}$ meter \(\begin{array}{ll} & \text { Current }=\text { IA } \\ \text { Let, } & \text { Length }=l\end{array}\) We know that, $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}$ $\mathrm{B}_{\mathrm{A}}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{b}}$ Force of wire due to field of A- $\mathrm{F}=\mathrm{IB}_{\mathrm{A}} \cdot l$ $\mathrm{~F}=\mathrm{I} \frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{b}} l$ $\frac{\mathrm{F}}{l}=\frac{\mu_{0} \mathrm{I}^{2}}{2 \pi \mathrm{b}} \mathrm{N} / \mathrm{m}$ Hence, magnitude of the force per unit length exerted by one wire on the other is $\frac{\mu_{0} I^{2}}{2 \pi b} \mathrm{~N} / \mathrm{m}$.
CG PET- 2004
Moving Charges & Magnetism
153800
A wire of length $\ell$ is bent to the form of a circular coil of some turns. A current i flows through the coil. The coil is placed in a uniform magnetic field $B$. The maximum torque on the coil can be
1 $\frac{\mathrm{iB} \ell^{2}}{4 \pi}$
2 $\frac{\mathrm{iB} \ell^{2}}{\pi}$
3 $\frac{\mathrm{iB} \ell^{2}}{2 \pi}$
4 $\frac{2 \mathrm{iB} \ell^{2}}{\pi}$
Explanation:
A We know that, $l=2 \pi \mathrm{RN}$ Where, $\mathrm{N}=\text { Number of turns }$ $\mathrm{R}=\text { Radius of the coil }$ $\mathrm{R}=\frac{l}{2 \pi \mathrm{N}}$ Magnetic moment $\mathrm{M}=\mathrm{NiA}=\mathrm{Ni}\left(\pi \mathrm{R}^{2}\right)$ $=\mathrm{Ni}\left(\frac{\pi l^{2}}{4 \pi^{2} \mathrm{~N}^{2}}\right)=\frac{\mathrm{i} l^{2}}{4 \pi \mathrm{N}}$ Maximum value of $\mathrm{M}$ can be, $\mathrm{M}_{\text {max }}=\frac{\mathrm{i} l^{2}}{4 \pi} \quad(\because \mathrm{N}=1)$ $\tau_{\max }=\mathrm{M}_{\max } \mathrm{B} \sin 90^{\circ}=\frac{\mathrm{iB} l^{2}}{4 \pi}$
BITSAT-2015
Moving Charges & Magnetism
153801
Two parallel wires carrying currents $I_{1}$ and $I_{2}$ in opposite directions and separated by a distance $d$ experience a
1 Repulsive force $\mu_{0} I_{1} I_{2} / 2 \pi d$
2 Attractive force $\mu_{0} I_{1} I_{2} / 2 \pi d$
3 Repulsive force $\mu_{0} \mathrm{I}_{1} \mathrm{I}_{2} / 2 \pi \mathrm{d}^{2}$
4 Attractive force $\mu_{0} I_{1} I_{2} / 2 \pi d^{2}$
Explanation:
A When two wires carrying currents $\mathrm{I}_{1} \& \mathrm{I}_{2}$ in opposite direction then they will repel each other. Magnetic field created by current in (1) wire. $\mathrm{B}_{1}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \pi \mathrm{d}}$ Now magnetic force on wire (2) due to wire (1) per unit length $\mathrm{F}_{21}=\mathrm{I}_{2} l \mathrm{~B}_{1}$ $\mathrm{~F}_{21}=\mathrm{I}_{2} l \times \frac{\mu_{0} \mathrm{I}_{1}}{2 \pi \mathrm{d}}$ $\frac{\mathrm{F}_{21}}{l}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1} \mathrm{I}_{2}}{2 \pi \mathrm{d}}$ $\mathrm{F}_{21}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1} \mathrm{I}_{2}}{2 \pi \mathrm{d}} \quad(\therefore l=1)$ Then, $\quad F_{21}=F_{12}=F=\frac{\mu_{0} I_{1} I_{2}}{2 \pi d}$ As the direction of currents are opposite, the force is repulsive force
UPSEE - 2011
Moving Charges & Magnetism
153802
The wire of length $l$ is bent into a circular loop of a single turn and is suspended in a magnetic field of induction $B$. When a current $I$ is passed through the loop, the maximum torque experienced by it is
A Let the radius of circular loop be $r$ then circumference of circular loop $l=2 \pi \mathrm{r} \quad \text { or } \quad \mathrm{r}=\frac{l}{2 \pi}$ When current $I$ is passed through the loop, the maximum torque acting on the loop is given by $\tau =\mathrm{I} \mathrm{B} \mathrm{A}$ $\tau =\mathrm{I} \mathrm{B} \times \pi \mathrm{r}^{2}$ $\tau =\mathrm{IB} \pi \times\left(\frac{l}{2 \pi}\right)^{2} \quad\left(\because \mathrm{r}=\frac{l}{2 \pi}\right)$ $\tau =\mathrm{IB} \pi \times \frac{l^{2}}{4 \pi^{2}}$ $\tau =\left(\frac{1}{4 \pi}\right) \mathrm{BI} l^{2}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Moving Charges & Magnetism
153799
Two thin long parallel wires separated by a distance $b$ meter are carrying a current of $I$ ampere each. The magnitude of the force per unit length exerted by one wire on the other is [ $\mu_{0}=$ permeability constant]
C Given, Distance between two wire $=\mathrm{b}$ meter \(\begin{array}{ll} & \text { Current }=\text { IA } \\ \text { Let, } & \text { Length }=l\end{array}\) We know that, $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}$ $\mathrm{B}_{\mathrm{A}}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{b}}$ Force of wire due to field of A- $\mathrm{F}=\mathrm{IB}_{\mathrm{A}} \cdot l$ $\mathrm{~F}=\mathrm{I} \frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{b}} l$ $\frac{\mathrm{F}}{l}=\frac{\mu_{0} \mathrm{I}^{2}}{2 \pi \mathrm{b}} \mathrm{N} / \mathrm{m}$ Hence, magnitude of the force per unit length exerted by one wire on the other is $\frac{\mu_{0} I^{2}}{2 \pi b} \mathrm{~N} / \mathrm{m}$.
CG PET- 2004
Moving Charges & Magnetism
153800
A wire of length $\ell$ is bent to the form of a circular coil of some turns. A current i flows through the coil. The coil is placed in a uniform magnetic field $B$. The maximum torque on the coil can be
1 $\frac{\mathrm{iB} \ell^{2}}{4 \pi}$
2 $\frac{\mathrm{iB} \ell^{2}}{\pi}$
3 $\frac{\mathrm{iB} \ell^{2}}{2 \pi}$
4 $\frac{2 \mathrm{iB} \ell^{2}}{\pi}$
Explanation:
A We know that, $l=2 \pi \mathrm{RN}$ Where, $\mathrm{N}=\text { Number of turns }$ $\mathrm{R}=\text { Radius of the coil }$ $\mathrm{R}=\frac{l}{2 \pi \mathrm{N}}$ Magnetic moment $\mathrm{M}=\mathrm{NiA}=\mathrm{Ni}\left(\pi \mathrm{R}^{2}\right)$ $=\mathrm{Ni}\left(\frac{\pi l^{2}}{4 \pi^{2} \mathrm{~N}^{2}}\right)=\frac{\mathrm{i} l^{2}}{4 \pi \mathrm{N}}$ Maximum value of $\mathrm{M}$ can be, $\mathrm{M}_{\text {max }}=\frac{\mathrm{i} l^{2}}{4 \pi} \quad(\because \mathrm{N}=1)$ $\tau_{\max }=\mathrm{M}_{\max } \mathrm{B} \sin 90^{\circ}=\frac{\mathrm{iB} l^{2}}{4 \pi}$
BITSAT-2015
Moving Charges & Magnetism
153801
Two parallel wires carrying currents $I_{1}$ and $I_{2}$ in opposite directions and separated by a distance $d$ experience a
1 Repulsive force $\mu_{0} I_{1} I_{2} / 2 \pi d$
2 Attractive force $\mu_{0} I_{1} I_{2} / 2 \pi d$
3 Repulsive force $\mu_{0} \mathrm{I}_{1} \mathrm{I}_{2} / 2 \pi \mathrm{d}^{2}$
4 Attractive force $\mu_{0} I_{1} I_{2} / 2 \pi d^{2}$
Explanation:
A When two wires carrying currents $\mathrm{I}_{1} \& \mathrm{I}_{2}$ in opposite direction then they will repel each other. Magnetic field created by current in (1) wire. $\mathrm{B}_{1}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \pi \mathrm{d}}$ Now magnetic force on wire (2) due to wire (1) per unit length $\mathrm{F}_{21}=\mathrm{I}_{2} l \mathrm{~B}_{1}$ $\mathrm{~F}_{21}=\mathrm{I}_{2} l \times \frac{\mu_{0} \mathrm{I}_{1}}{2 \pi \mathrm{d}}$ $\frac{\mathrm{F}_{21}}{l}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1} \mathrm{I}_{2}}{2 \pi \mathrm{d}}$ $\mathrm{F}_{21}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1} \mathrm{I}_{2}}{2 \pi \mathrm{d}} \quad(\therefore l=1)$ Then, $\quad F_{21}=F_{12}=F=\frac{\mu_{0} I_{1} I_{2}}{2 \pi d}$ As the direction of currents are opposite, the force is repulsive force
UPSEE - 2011
Moving Charges & Magnetism
153802
The wire of length $l$ is bent into a circular loop of a single turn and is suspended in a magnetic field of induction $B$. When a current $I$ is passed through the loop, the maximum torque experienced by it is
A Let the radius of circular loop be $r$ then circumference of circular loop $l=2 \pi \mathrm{r} \quad \text { or } \quad \mathrm{r}=\frac{l}{2 \pi}$ When current $I$ is passed through the loop, the maximum torque acting on the loop is given by $\tau =\mathrm{I} \mathrm{B} \mathrm{A}$ $\tau =\mathrm{I} \mathrm{B} \times \pi \mathrm{r}^{2}$ $\tau =\mathrm{IB} \pi \times\left(\frac{l}{2 \pi}\right)^{2} \quad\left(\because \mathrm{r}=\frac{l}{2 \pi}\right)$ $\tau =\mathrm{IB} \pi \times \frac{l^{2}}{4 \pi^{2}}$ $\tau =\left(\frac{1}{4 \pi}\right) \mathrm{BI} l^{2}$
