153440
Three parallel wires carrying currents are as shown in the figure. The directions of net force on wire ' $b$ ' due to wire ' $a$ ' and ' $c$ ' is
1 along the direction of current through wire ' $b$ '
2 towards ' $\mathrm{c}$ '
3 towards 'a'
4 opposite to the direction of current through wire ' $b$ '
Explanation:
C Wire a and b carry current in the same direction then they will attract each other. So force on $b$ due to a will be towards a wire b and c carry currents in opposite direction. The force on $\mathrm{b}$ due to $\mathrm{c}$ will be towards a hence the net force will be towards a.
MHT-CET 2020
Moving Charges & Magnetism
153446
A toroid has an iron core with an internal magnetic field of $10 \pi \mathrm{mT}$, when the current in the winding of 1500 turns per meter is $10 \mathrm{~A}$. Determine the field due to magnetisation $\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{Hm}^{-1}\right)$
1 $(4 \pi) \mathrm{mT}$
2 $(10 \pi) \mathrm{mT}$
3 $\left(\frac{8}{\pi}\right) \mathrm{mT}$
4 $\left(\frac{\pi}{4}\right) \mathrm{mT}$
Explanation:
A Given, $\mathrm{B}=10 \pi \mathrm{mT}$ $\mathrm{n}=1500$ turn $\mathrm{I}=10 \mathrm{~A}$ $\mu_{0}=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}$ Magnetic field of toroid $\mathrm{B}_{\mathrm{t}} =\mu_{0} \mathrm{nI}$ $=4 \pi \times 10^{-7} \times 1500 \times 10$ $\mathrm{~B}_{\mathrm{t}} =6 \pi \mathrm{mT}$ Hence the field due to magnetisation $\mathrm{B}_{\mathrm{m}} =\mathrm{B}-\mathrm{B}_{\mathrm{t}}$ $=10 \pi \mathrm{mT}-6 \pi \mathrm{mT}$ $\mathrm{B}_{\mathrm{m}} =4 \pi \mathrm{mT}$
TS- EAMCET-03.05.2019
Moving Charges & Magnetism
153448
Four identical long solenoids A, B, C, D are connected as in figure. If magnetic field at the centre of $A$ is $3.0 \mathrm{~T}$, the field at the centre of $B$ will be
1 $3.0 \mathrm{~T}$
2 $6.0 \mathrm{~T}$
3 $1.5 \mathrm{~T}$
4 $12.0 \mathrm{~T}$
Explanation:
C Given, magnetic field at the centre of $\mathrm{A}$ solenoids, $\mathrm{B}_{\mathrm{A}}=3 \mathrm{~T}$ Let, the current I flow in solenoids A. Then this current is equally distributed in solenoid $\mathrm{C}$ and $\mathrm{B}$. We know that, $\mathrm{B}=\mu_{0} \mathrm{nI}$ $\mathrm{B} \propto \mathrm{I}$ $\frac{\mathrm{B}_{\mathrm{B}}}{\mathrm{B}_{\mathrm{A}}}=\frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{A}}}$ $\mathrm{B}_{\mathrm{B}}=\frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{A}}} \times \mathrm{B}_{\mathrm{A}}=\frac{\mathrm{I} / 2}{\mathrm{I}} \times 3 \mathrm{~T}$ $\mathrm{~B}_{\mathrm{B}}=\frac{3}{2} \mathrm{~T}=1.5 \mathrm{~T}$
Assam CEE-2019
Moving Charges & Magnetism
153449
A solenoid with 500 turns of wire is $25 \mathrm{~cm}$ long and is filled with magnetic material of susceptibility $\chi_{\mathrm{m}}$ is (Assume $\mu_{0}=4 \pi \times 10^{-7}$ $\mathbf{T m} / \mathbf{A})$
1 499
2 550
3 350
4 649
Explanation:
A Given, $\mathrm{N}=500$ turns, $l=25 \mathrm{~cm}=0.25 \mathrm{~m}$ Current $(\mathrm{I})=2 \mathrm{~A}$ Magnetic field $(B)=(0.8 \pi) \mathrm{T}$ Magnetic field, $\mathrm{B}=\mu_{0} \mu_{\mathrm{r}} \mathrm{nI}=\mu_{0} \mu_{\mathrm{r}} \frac{\mathrm{N}}{l} \mathrm{I}$ $0.8 \pi=4 \pi \times 10^{-7} \times \mu_{\mathrm{r}} \times \frac{500}{0.25} \times 2$ $\mu_{\mathrm{r}}=\frac{0.8 \pi \times 0.25}{4 \pi \times 10^{-7} \times 500 \times 2}=500$ $\mu_{\mathrm{r}}=1+\chi_{\mathrm{m}}$ $\chi_{\mathrm{m}}=\mu_{\mathrm{r}}-1=500-1=499$
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Moving Charges & Magnetism
153440
Three parallel wires carrying currents are as shown in the figure. The directions of net force on wire ' $b$ ' due to wire ' $a$ ' and ' $c$ ' is
1 along the direction of current through wire ' $b$ '
2 towards ' $\mathrm{c}$ '
3 towards 'a'
4 opposite to the direction of current through wire ' $b$ '
Explanation:
C Wire a and b carry current in the same direction then they will attract each other. So force on $b$ due to a will be towards a wire b and c carry currents in opposite direction. The force on $\mathrm{b}$ due to $\mathrm{c}$ will be towards a hence the net force will be towards a.
MHT-CET 2020
Moving Charges & Magnetism
153446
A toroid has an iron core with an internal magnetic field of $10 \pi \mathrm{mT}$, when the current in the winding of 1500 turns per meter is $10 \mathrm{~A}$. Determine the field due to magnetisation $\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{Hm}^{-1}\right)$
1 $(4 \pi) \mathrm{mT}$
2 $(10 \pi) \mathrm{mT}$
3 $\left(\frac{8}{\pi}\right) \mathrm{mT}$
4 $\left(\frac{\pi}{4}\right) \mathrm{mT}$
Explanation:
A Given, $\mathrm{B}=10 \pi \mathrm{mT}$ $\mathrm{n}=1500$ turn $\mathrm{I}=10 \mathrm{~A}$ $\mu_{0}=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}$ Magnetic field of toroid $\mathrm{B}_{\mathrm{t}} =\mu_{0} \mathrm{nI}$ $=4 \pi \times 10^{-7} \times 1500 \times 10$ $\mathrm{~B}_{\mathrm{t}} =6 \pi \mathrm{mT}$ Hence the field due to magnetisation $\mathrm{B}_{\mathrm{m}} =\mathrm{B}-\mathrm{B}_{\mathrm{t}}$ $=10 \pi \mathrm{mT}-6 \pi \mathrm{mT}$ $\mathrm{B}_{\mathrm{m}} =4 \pi \mathrm{mT}$
TS- EAMCET-03.05.2019
Moving Charges & Magnetism
153448
Four identical long solenoids A, B, C, D are connected as in figure. If magnetic field at the centre of $A$ is $3.0 \mathrm{~T}$, the field at the centre of $B$ will be
1 $3.0 \mathrm{~T}$
2 $6.0 \mathrm{~T}$
3 $1.5 \mathrm{~T}$
4 $12.0 \mathrm{~T}$
Explanation:
C Given, magnetic field at the centre of $\mathrm{A}$ solenoids, $\mathrm{B}_{\mathrm{A}}=3 \mathrm{~T}$ Let, the current I flow in solenoids A. Then this current is equally distributed in solenoid $\mathrm{C}$ and $\mathrm{B}$. We know that, $\mathrm{B}=\mu_{0} \mathrm{nI}$ $\mathrm{B} \propto \mathrm{I}$ $\frac{\mathrm{B}_{\mathrm{B}}}{\mathrm{B}_{\mathrm{A}}}=\frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{A}}}$ $\mathrm{B}_{\mathrm{B}}=\frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{A}}} \times \mathrm{B}_{\mathrm{A}}=\frac{\mathrm{I} / 2}{\mathrm{I}} \times 3 \mathrm{~T}$ $\mathrm{~B}_{\mathrm{B}}=\frac{3}{2} \mathrm{~T}=1.5 \mathrm{~T}$
Assam CEE-2019
Moving Charges & Magnetism
153449
A solenoid with 500 turns of wire is $25 \mathrm{~cm}$ long and is filled with magnetic material of susceptibility $\chi_{\mathrm{m}}$ is (Assume $\mu_{0}=4 \pi \times 10^{-7}$ $\mathbf{T m} / \mathbf{A})$
1 499
2 550
3 350
4 649
Explanation:
A Given, $\mathrm{N}=500$ turns, $l=25 \mathrm{~cm}=0.25 \mathrm{~m}$ Current $(\mathrm{I})=2 \mathrm{~A}$ Magnetic field $(B)=(0.8 \pi) \mathrm{T}$ Magnetic field, $\mathrm{B}=\mu_{0} \mu_{\mathrm{r}} \mathrm{nI}=\mu_{0} \mu_{\mathrm{r}} \frac{\mathrm{N}}{l} \mathrm{I}$ $0.8 \pi=4 \pi \times 10^{-7} \times \mu_{\mathrm{r}} \times \frac{500}{0.25} \times 2$ $\mu_{\mathrm{r}}=\frac{0.8 \pi \times 0.25}{4 \pi \times 10^{-7} \times 500 \times 2}=500$ $\mu_{\mathrm{r}}=1+\chi_{\mathrm{m}}$ $\chi_{\mathrm{m}}=\mu_{\mathrm{r}}-1=500-1=499$
153440
Three parallel wires carrying currents are as shown in the figure. The directions of net force on wire ' $b$ ' due to wire ' $a$ ' and ' $c$ ' is
1 along the direction of current through wire ' $b$ '
2 towards ' $\mathrm{c}$ '
3 towards 'a'
4 opposite to the direction of current through wire ' $b$ '
Explanation:
C Wire a and b carry current in the same direction then they will attract each other. So force on $b$ due to a will be towards a wire b and c carry currents in opposite direction. The force on $\mathrm{b}$ due to $\mathrm{c}$ will be towards a hence the net force will be towards a.
MHT-CET 2020
Moving Charges & Magnetism
153446
A toroid has an iron core with an internal magnetic field of $10 \pi \mathrm{mT}$, when the current in the winding of 1500 turns per meter is $10 \mathrm{~A}$. Determine the field due to magnetisation $\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{Hm}^{-1}\right)$
1 $(4 \pi) \mathrm{mT}$
2 $(10 \pi) \mathrm{mT}$
3 $\left(\frac{8}{\pi}\right) \mathrm{mT}$
4 $\left(\frac{\pi}{4}\right) \mathrm{mT}$
Explanation:
A Given, $\mathrm{B}=10 \pi \mathrm{mT}$ $\mathrm{n}=1500$ turn $\mathrm{I}=10 \mathrm{~A}$ $\mu_{0}=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}$ Magnetic field of toroid $\mathrm{B}_{\mathrm{t}} =\mu_{0} \mathrm{nI}$ $=4 \pi \times 10^{-7} \times 1500 \times 10$ $\mathrm{~B}_{\mathrm{t}} =6 \pi \mathrm{mT}$ Hence the field due to magnetisation $\mathrm{B}_{\mathrm{m}} =\mathrm{B}-\mathrm{B}_{\mathrm{t}}$ $=10 \pi \mathrm{mT}-6 \pi \mathrm{mT}$ $\mathrm{B}_{\mathrm{m}} =4 \pi \mathrm{mT}$
TS- EAMCET-03.05.2019
Moving Charges & Magnetism
153448
Four identical long solenoids A, B, C, D are connected as in figure. If magnetic field at the centre of $A$ is $3.0 \mathrm{~T}$, the field at the centre of $B$ will be
1 $3.0 \mathrm{~T}$
2 $6.0 \mathrm{~T}$
3 $1.5 \mathrm{~T}$
4 $12.0 \mathrm{~T}$
Explanation:
C Given, magnetic field at the centre of $\mathrm{A}$ solenoids, $\mathrm{B}_{\mathrm{A}}=3 \mathrm{~T}$ Let, the current I flow in solenoids A. Then this current is equally distributed in solenoid $\mathrm{C}$ and $\mathrm{B}$. We know that, $\mathrm{B}=\mu_{0} \mathrm{nI}$ $\mathrm{B} \propto \mathrm{I}$ $\frac{\mathrm{B}_{\mathrm{B}}}{\mathrm{B}_{\mathrm{A}}}=\frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{A}}}$ $\mathrm{B}_{\mathrm{B}}=\frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{A}}} \times \mathrm{B}_{\mathrm{A}}=\frac{\mathrm{I} / 2}{\mathrm{I}} \times 3 \mathrm{~T}$ $\mathrm{~B}_{\mathrm{B}}=\frac{3}{2} \mathrm{~T}=1.5 \mathrm{~T}$
Assam CEE-2019
Moving Charges & Magnetism
153449
A solenoid with 500 turns of wire is $25 \mathrm{~cm}$ long and is filled with magnetic material of susceptibility $\chi_{\mathrm{m}}$ is (Assume $\mu_{0}=4 \pi \times 10^{-7}$ $\mathbf{T m} / \mathbf{A})$
1 499
2 550
3 350
4 649
Explanation:
A Given, $\mathrm{N}=500$ turns, $l=25 \mathrm{~cm}=0.25 \mathrm{~m}$ Current $(\mathrm{I})=2 \mathrm{~A}$ Magnetic field $(B)=(0.8 \pi) \mathrm{T}$ Magnetic field, $\mathrm{B}=\mu_{0} \mu_{\mathrm{r}} \mathrm{nI}=\mu_{0} \mu_{\mathrm{r}} \frac{\mathrm{N}}{l} \mathrm{I}$ $0.8 \pi=4 \pi \times 10^{-7} \times \mu_{\mathrm{r}} \times \frac{500}{0.25} \times 2$ $\mu_{\mathrm{r}}=\frac{0.8 \pi \times 0.25}{4 \pi \times 10^{-7} \times 500 \times 2}=500$ $\mu_{\mathrm{r}}=1+\chi_{\mathrm{m}}$ $\chi_{\mathrm{m}}=\mu_{\mathrm{r}}-1=500-1=499$
153440
Three parallel wires carrying currents are as shown in the figure. The directions of net force on wire ' $b$ ' due to wire ' $a$ ' and ' $c$ ' is
1 along the direction of current through wire ' $b$ '
2 towards ' $\mathrm{c}$ '
3 towards 'a'
4 opposite to the direction of current through wire ' $b$ '
Explanation:
C Wire a and b carry current in the same direction then they will attract each other. So force on $b$ due to a will be towards a wire b and c carry currents in opposite direction. The force on $\mathrm{b}$ due to $\mathrm{c}$ will be towards a hence the net force will be towards a.
MHT-CET 2020
Moving Charges & Magnetism
153446
A toroid has an iron core with an internal magnetic field of $10 \pi \mathrm{mT}$, when the current in the winding of 1500 turns per meter is $10 \mathrm{~A}$. Determine the field due to magnetisation $\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{Hm}^{-1}\right)$
1 $(4 \pi) \mathrm{mT}$
2 $(10 \pi) \mathrm{mT}$
3 $\left(\frac{8}{\pi}\right) \mathrm{mT}$
4 $\left(\frac{\pi}{4}\right) \mathrm{mT}$
Explanation:
A Given, $\mathrm{B}=10 \pi \mathrm{mT}$ $\mathrm{n}=1500$ turn $\mathrm{I}=10 \mathrm{~A}$ $\mu_{0}=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}$ Magnetic field of toroid $\mathrm{B}_{\mathrm{t}} =\mu_{0} \mathrm{nI}$ $=4 \pi \times 10^{-7} \times 1500 \times 10$ $\mathrm{~B}_{\mathrm{t}} =6 \pi \mathrm{mT}$ Hence the field due to magnetisation $\mathrm{B}_{\mathrm{m}} =\mathrm{B}-\mathrm{B}_{\mathrm{t}}$ $=10 \pi \mathrm{mT}-6 \pi \mathrm{mT}$ $\mathrm{B}_{\mathrm{m}} =4 \pi \mathrm{mT}$
TS- EAMCET-03.05.2019
Moving Charges & Magnetism
153448
Four identical long solenoids A, B, C, D are connected as in figure. If magnetic field at the centre of $A$ is $3.0 \mathrm{~T}$, the field at the centre of $B$ will be
1 $3.0 \mathrm{~T}$
2 $6.0 \mathrm{~T}$
3 $1.5 \mathrm{~T}$
4 $12.0 \mathrm{~T}$
Explanation:
C Given, magnetic field at the centre of $\mathrm{A}$ solenoids, $\mathrm{B}_{\mathrm{A}}=3 \mathrm{~T}$ Let, the current I flow in solenoids A. Then this current is equally distributed in solenoid $\mathrm{C}$ and $\mathrm{B}$. We know that, $\mathrm{B}=\mu_{0} \mathrm{nI}$ $\mathrm{B} \propto \mathrm{I}$ $\frac{\mathrm{B}_{\mathrm{B}}}{\mathrm{B}_{\mathrm{A}}}=\frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{A}}}$ $\mathrm{B}_{\mathrm{B}}=\frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{A}}} \times \mathrm{B}_{\mathrm{A}}=\frac{\mathrm{I} / 2}{\mathrm{I}} \times 3 \mathrm{~T}$ $\mathrm{~B}_{\mathrm{B}}=\frac{3}{2} \mathrm{~T}=1.5 \mathrm{~T}$
Assam CEE-2019
Moving Charges & Magnetism
153449
A solenoid with 500 turns of wire is $25 \mathrm{~cm}$ long and is filled with magnetic material of susceptibility $\chi_{\mathrm{m}}$ is (Assume $\mu_{0}=4 \pi \times 10^{-7}$ $\mathbf{T m} / \mathbf{A})$
1 499
2 550
3 350
4 649
Explanation:
A Given, $\mathrm{N}=500$ turns, $l=25 \mathrm{~cm}=0.25 \mathrm{~m}$ Current $(\mathrm{I})=2 \mathrm{~A}$ Magnetic field $(B)=(0.8 \pi) \mathrm{T}$ Magnetic field, $\mathrm{B}=\mu_{0} \mu_{\mathrm{r}} \mathrm{nI}=\mu_{0} \mu_{\mathrm{r}} \frac{\mathrm{N}}{l} \mathrm{I}$ $0.8 \pi=4 \pi \times 10^{-7} \times \mu_{\mathrm{r}} \times \frac{500}{0.25} \times 2$ $\mu_{\mathrm{r}}=\frac{0.8 \pi \times 0.25}{4 \pi \times 10^{-7} \times 500 \times 2}=500$ $\mu_{\mathrm{r}}=1+\chi_{\mathrm{m}}$ $\chi_{\mathrm{m}}=\mu_{\mathrm{r}}-1=500-1=499$
