153259
1A current flows through an infinitely long straight wire. The magnetic field produced at a point $1 \mathrm{~m}$ away from it is
1 $2 \times 10^{-3} \mathrm{~T}$
2 $2 / 10 \mathrm{~T}$
3 $2 \times 10^{-7} \mathrm{~T}$
4 $2 \pi \times 10^{-6} \mathrm{~T}$
Explanation:
C According to the question, Magnetic field due infinite long wire \(B=\frac{\mu_0}{2 \pi} \frac{I}{r}\) \(\text { Given, } r =1 \mathrm{~m}\) \(I =1 \mathrm{~A}\) \(\therefore \quad B =\frac{4 \pi \times 10^{-7}}{2 \pi} \times \frac{1}{1} \quad\left(\because \mu_0=4 \pi \times 10^{-7}\right)\) \(B=2 \times 10^{-7} \mathrm{~T}\)
CG PET- 2004
Moving Charges & Magnetism
153260
A particle is moving with velocity $\vec{v}=\hat{i}+3 \hat{j}$ and it produces an electric field at a point given by $\overrightarrow{\mathbf{E}}=\mathbf{2} \hat{\mathbf{k}}$. It will produce magnetic field at that point equal to (all quantities are in SI units)
1 $\frac{6 \hat{i}-2 \hat{j}}{c^{2}}$
2 $\frac{6 \hat{i}+2 \hat{j}}{c^{2}}$
3 zero
4 cannot be determined from the given data
Explanation:
A Electric field $\mathrm{E}$ $\overrightarrow{\mathrm{E}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}}$ Given $\vec{E}=2 \hat{k}$, velocity $\vec{v}=\hat{i}+3 \hat{j}$ $\therefore \quad \frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}}=2 \hat{\mathrm{k}}, \quad(\hat{\mathrm{r}}=\hat{\mathrm{k}})$ Magnetic field in vector form, $\overrightarrow{\mathrm{B}}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{q}(\overrightarrow{\mathrm{v}} \times \hat{\mathrm{r}})}{\mathrm{r}^{2}}$ $\overrightarrow{\mathrm{B}}=\mu_{0} \varepsilon_{0}\left[\frac{\mathrm{q}}{4 \pi \varepsilon_{0} \mathrm{r}^{2}}\right](\hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \times \hat{\mathrm{k}}$ $\overrightarrow{\mathrm{B}}=\mu_{0} \varepsilon_{0} \times 2[-\hat{\mathrm{j}}+3 \hat{\mathrm{i}}]$ $\overrightarrow{\mathrm{B}}=\mu_{0} \varepsilon_{0}(6 \hat{\mathrm{i}}-2 \hat{\mathrm{j}})$ $\mathrm{c}=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$ $\mu_{0} \varepsilon_{0}=\frac{1}{\mathrm{c}^{2}}$ $\because \quad \mathrm{c}=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$ Putting the value of $\mu_{0} \varepsilon_{0}$ in equation (i), we get- $\therefore \quad \overrightarrow{\mathrm{B}}=\frac{(6 \hat{\mathrm{i}}-2 \hat{\mathrm{j}})}{\mathrm{c}^{2}}$
BITSAT-2005
Moving Charges & Magnetism
153261
Calculate the magnetic field at the centre of a coil in the form of a square of side 2a carrying a current $I$.
B According to the question, We know that, side of square $=2 \mathrm{a}$ Magnetic field $(B)=\frac{\mu_{0}}{4 \pi \mathrm{r}} \mathrm{I}(\sin \alpha+\sin \beta)$ From the figure, $\alpha= \beta=45^{\circ}$ $\mathrm{B}_{\text {net }} =4 \times \frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{a}}\left[\sin 45+\sin 45^{\circ}\right]$ $\mathrm{B}_{\text {net }} =4 \times \frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{a}}\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right]$ $\mathrm{B}_{\text {net }} =\frac{4 \mu_{0} \mathrm{I}}{4 \pi \mathrm{a}} \frac{2}{\sqrt{2}}$ $\mathrm{~B}_{\text {net }} =\frac{4 \mu_{0} \mathrm{I}}{4 \pi \mathrm{a}} \times \frac{2 \sqrt{2}}{2}$ $\mathrm{~B}_{\text {net }} =\frac{\sqrt{2} \mu_{0} \mathrm{I}}{\pi \mathrm{a}}$
AIIMS-2012
Moving Charges & Magnetism
153262
A steady current is set up in a cubic network composed of wires of equal resistance and length $d$ as shown in figure. What is the magnetic field at the centre $P$ due to the cubic network
C From the given figure, All the diagonally opposite arms will carry same current. So, the net magnetic field at $\mathrm{P}$ will be zero. Because the magnetic field at center $\mathrm{P}$ is cancelled by opposite pairs. Ex. AD \& EF will cancel each other. Therefore, magnetic field at the centre $\mathrm{P}$ due to the cubic network is zero.
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Moving Charges & Magnetism
153259
1A current flows through an infinitely long straight wire. The magnetic field produced at a point $1 \mathrm{~m}$ away from it is
1 $2 \times 10^{-3} \mathrm{~T}$
2 $2 / 10 \mathrm{~T}$
3 $2 \times 10^{-7} \mathrm{~T}$
4 $2 \pi \times 10^{-6} \mathrm{~T}$
Explanation:
C According to the question, Magnetic field due infinite long wire \(B=\frac{\mu_0}{2 \pi} \frac{I}{r}\) \(\text { Given, } r =1 \mathrm{~m}\) \(I =1 \mathrm{~A}\) \(\therefore \quad B =\frac{4 \pi \times 10^{-7}}{2 \pi} \times \frac{1}{1} \quad\left(\because \mu_0=4 \pi \times 10^{-7}\right)\) \(B=2 \times 10^{-7} \mathrm{~T}\)
CG PET- 2004
Moving Charges & Magnetism
153260
A particle is moving with velocity $\vec{v}=\hat{i}+3 \hat{j}$ and it produces an electric field at a point given by $\overrightarrow{\mathbf{E}}=\mathbf{2} \hat{\mathbf{k}}$. It will produce magnetic field at that point equal to (all quantities are in SI units)
1 $\frac{6 \hat{i}-2 \hat{j}}{c^{2}}$
2 $\frac{6 \hat{i}+2 \hat{j}}{c^{2}}$
3 zero
4 cannot be determined from the given data
Explanation:
A Electric field $\mathrm{E}$ $\overrightarrow{\mathrm{E}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}}$ Given $\vec{E}=2 \hat{k}$, velocity $\vec{v}=\hat{i}+3 \hat{j}$ $\therefore \quad \frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}}=2 \hat{\mathrm{k}}, \quad(\hat{\mathrm{r}}=\hat{\mathrm{k}})$ Magnetic field in vector form, $\overrightarrow{\mathrm{B}}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{q}(\overrightarrow{\mathrm{v}} \times \hat{\mathrm{r}})}{\mathrm{r}^{2}}$ $\overrightarrow{\mathrm{B}}=\mu_{0} \varepsilon_{0}\left[\frac{\mathrm{q}}{4 \pi \varepsilon_{0} \mathrm{r}^{2}}\right](\hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \times \hat{\mathrm{k}}$ $\overrightarrow{\mathrm{B}}=\mu_{0} \varepsilon_{0} \times 2[-\hat{\mathrm{j}}+3 \hat{\mathrm{i}}]$ $\overrightarrow{\mathrm{B}}=\mu_{0} \varepsilon_{0}(6 \hat{\mathrm{i}}-2 \hat{\mathrm{j}})$ $\mathrm{c}=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$ $\mu_{0} \varepsilon_{0}=\frac{1}{\mathrm{c}^{2}}$ $\because \quad \mathrm{c}=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$ Putting the value of $\mu_{0} \varepsilon_{0}$ in equation (i), we get- $\therefore \quad \overrightarrow{\mathrm{B}}=\frac{(6 \hat{\mathrm{i}}-2 \hat{\mathrm{j}})}{\mathrm{c}^{2}}$
BITSAT-2005
Moving Charges & Magnetism
153261
Calculate the magnetic field at the centre of a coil in the form of a square of side 2a carrying a current $I$.
B According to the question, We know that, side of square $=2 \mathrm{a}$ Magnetic field $(B)=\frac{\mu_{0}}{4 \pi \mathrm{r}} \mathrm{I}(\sin \alpha+\sin \beta)$ From the figure, $\alpha= \beta=45^{\circ}$ $\mathrm{B}_{\text {net }} =4 \times \frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{a}}\left[\sin 45+\sin 45^{\circ}\right]$ $\mathrm{B}_{\text {net }} =4 \times \frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{a}}\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right]$ $\mathrm{B}_{\text {net }} =\frac{4 \mu_{0} \mathrm{I}}{4 \pi \mathrm{a}} \frac{2}{\sqrt{2}}$ $\mathrm{~B}_{\text {net }} =\frac{4 \mu_{0} \mathrm{I}}{4 \pi \mathrm{a}} \times \frac{2 \sqrt{2}}{2}$ $\mathrm{~B}_{\text {net }} =\frac{\sqrt{2} \mu_{0} \mathrm{I}}{\pi \mathrm{a}}$
AIIMS-2012
Moving Charges & Magnetism
153262
A steady current is set up in a cubic network composed of wires of equal resistance and length $d$ as shown in figure. What is the magnetic field at the centre $P$ due to the cubic network
C From the given figure, All the diagonally opposite arms will carry same current. So, the net magnetic field at $\mathrm{P}$ will be zero. Because the magnetic field at center $\mathrm{P}$ is cancelled by opposite pairs. Ex. AD \& EF will cancel each other. Therefore, magnetic field at the centre $\mathrm{P}$ due to the cubic network is zero.
153259
1A current flows through an infinitely long straight wire. The magnetic field produced at a point $1 \mathrm{~m}$ away from it is
1 $2 \times 10^{-3} \mathrm{~T}$
2 $2 / 10 \mathrm{~T}$
3 $2 \times 10^{-7} \mathrm{~T}$
4 $2 \pi \times 10^{-6} \mathrm{~T}$
Explanation:
C According to the question, Magnetic field due infinite long wire \(B=\frac{\mu_0}{2 \pi} \frac{I}{r}\) \(\text { Given, } r =1 \mathrm{~m}\) \(I =1 \mathrm{~A}\) \(\therefore \quad B =\frac{4 \pi \times 10^{-7}}{2 \pi} \times \frac{1}{1} \quad\left(\because \mu_0=4 \pi \times 10^{-7}\right)\) \(B=2 \times 10^{-7} \mathrm{~T}\)
CG PET- 2004
Moving Charges & Magnetism
153260
A particle is moving with velocity $\vec{v}=\hat{i}+3 \hat{j}$ and it produces an electric field at a point given by $\overrightarrow{\mathbf{E}}=\mathbf{2} \hat{\mathbf{k}}$. It will produce magnetic field at that point equal to (all quantities are in SI units)
1 $\frac{6 \hat{i}-2 \hat{j}}{c^{2}}$
2 $\frac{6 \hat{i}+2 \hat{j}}{c^{2}}$
3 zero
4 cannot be determined from the given data
Explanation:
A Electric field $\mathrm{E}$ $\overrightarrow{\mathrm{E}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}}$ Given $\vec{E}=2 \hat{k}$, velocity $\vec{v}=\hat{i}+3 \hat{j}$ $\therefore \quad \frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}}=2 \hat{\mathrm{k}}, \quad(\hat{\mathrm{r}}=\hat{\mathrm{k}})$ Magnetic field in vector form, $\overrightarrow{\mathrm{B}}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{q}(\overrightarrow{\mathrm{v}} \times \hat{\mathrm{r}})}{\mathrm{r}^{2}}$ $\overrightarrow{\mathrm{B}}=\mu_{0} \varepsilon_{0}\left[\frac{\mathrm{q}}{4 \pi \varepsilon_{0} \mathrm{r}^{2}}\right](\hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \times \hat{\mathrm{k}}$ $\overrightarrow{\mathrm{B}}=\mu_{0} \varepsilon_{0} \times 2[-\hat{\mathrm{j}}+3 \hat{\mathrm{i}}]$ $\overrightarrow{\mathrm{B}}=\mu_{0} \varepsilon_{0}(6 \hat{\mathrm{i}}-2 \hat{\mathrm{j}})$ $\mathrm{c}=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$ $\mu_{0} \varepsilon_{0}=\frac{1}{\mathrm{c}^{2}}$ $\because \quad \mathrm{c}=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$ Putting the value of $\mu_{0} \varepsilon_{0}$ in equation (i), we get- $\therefore \quad \overrightarrow{\mathrm{B}}=\frac{(6 \hat{\mathrm{i}}-2 \hat{\mathrm{j}})}{\mathrm{c}^{2}}$
BITSAT-2005
Moving Charges & Magnetism
153261
Calculate the magnetic field at the centre of a coil in the form of a square of side 2a carrying a current $I$.
B According to the question, We know that, side of square $=2 \mathrm{a}$ Magnetic field $(B)=\frac{\mu_{0}}{4 \pi \mathrm{r}} \mathrm{I}(\sin \alpha+\sin \beta)$ From the figure, $\alpha= \beta=45^{\circ}$ $\mathrm{B}_{\text {net }} =4 \times \frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{a}}\left[\sin 45+\sin 45^{\circ}\right]$ $\mathrm{B}_{\text {net }} =4 \times \frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{a}}\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right]$ $\mathrm{B}_{\text {net }} =\frac{4 \mu_{0} \mathrm{I}}{4 \pi \mathrm{a}} \frac{2}{\sqrt{2}}$ $\mathrm{~B}_{\text {net }} =\frac{4 \mu_{0} \mathrm{I}}{4 \pi \mathrm{a}} \times \frac{2 \sqrt{2}}{2}$ $\mathrm{~B}_{\text {net }} =\frac{\sqrt{2} \mu_{0} \mathrm{I}}{\pi \mathrm{a}}$
AIIMS-2012
Moving Charges & Magnetism
153262
A steady current is set up in a cubic network composed of wires of equal resistance and length $d$ as shown in figure. What is the magnetic field at the centre $P$ due to the cubic network
C From the given figure, All the diagonally opposite arms will carry same current. So, the net magnetic field at $\mathrm{P}$ will be zero. Because the magnetic field at center $\mathrm{P}$ is cancelled by opposite pairs. Ex. AD \& EF will cancel each other. Therefore, magnetic field at the centre $\mathrm{P}$ due to the cubic network is zero.
153259
1A current flows through an infinitely long straight wire. The magnetic field produced at a point $1 \mathrm{~m}$ away from it is
1 $2 \times 10^{-3} \mathrm{~T}$
2 $2 / 10 \mathrm{~T}$
3 $2 \times 10^{-7} \mathrm{~T}$
4 $2 \pi \times 10^{-6} \mathrm{~T}$
Explanation:
C According to the question, Magnetic field due infinite long wire \(B=\frac{\mu_0}{2 \pi} \frac{I}{r}\) \(\text { Given, } r =1 \mathrm{~m}\) \(I =1 \mathrm{~A}\) \(\therefore \quad B =\frac{4 \pi \times 10^{-7}}{2 \pi} \times \frac{1}{1} \quad\left(\because \mu_0=4 \pi \times 10^{-7}\right)\) \(B=2 \times 10^{-7} \mathrm{~T}\)
CG PET- 2004
Moving Charges & Magnetism
153260
A particle is moving with velocity $\vec{v}=\hat{i}+3 \hat{j}$ and it produces an electric field at a point given by $\overrightarrow{\mathbf{E}}=\mathbf{2} \hat{\mathbf{k}}$. It will produce magnetic field at that point equal to (all quantities are in SI units)
1 $\frac{6 \hat{i}-2 \hat{j}}{c^{2}}$
2 $\frac{6 \hat{i}+2 \hat{j}}{c^{2}}$
3 zero
4 cannot be determined from the given data
Explanation:
A Electric field $\mathrm{E}$ $\overrightarrow{\mathrm{E}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}}$ Given $\vec{E}=2 \hat{k}$, velocity $\vec{v}=\hat{i}+3 \hat{j}$ $\therefore \quad \frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}}=2 \hat{\mathrm{k}}, \quad(\hat{\mathrm{r}}=\hat{\mathrm{k}})$ Magnetic field in vector form, $\overrightarrow{\mathrm{B}}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{q}(\overrightarrow{\mathrm{v}} \times \hat{\mathrm{r}})}{\mathrm{r}^{2}}$ $\overrightarrow{\mathrm{B}}=\mu_{0} \varepsilon_{0}\left[\frac{\mathrm{q}}{4 \pi \varepsilon_{0} \mathrm{r}^{2}}\right](\hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \times \hat{\mathrm{k}}$ $\overrightarrow{\mathrm{B}}=\mu_{0} \varepsilon_{0} \times 2[-\hat{\mathrm{j}}+3 \hat{\mathrm{i}}]$ $\overrightarrow{\mathrm{B}}=\mu_{0} \varepsilon_{0}(6 \hat{\mathrm{i}}-2 \hat{\mathrm{j}})$ $\mathrm{c}=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$ $\mu_{0} \varepsilon_{0}=\frac{1}{\mathrm{c}^{2}}$ $\because \quad \mathrm{c}=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$ Putting the value of $\mu_{0} \varepsilon_{0}$ in equation (i), we get- $\therefore \quad \overrightarrow{\mathrm{B}}=\frac{(6 \hat{\mathrm{i}}-2 \hat{\mathrm{j}})}{\mathrm{c}^{2}}$
BITSAT-2005
Moving Charges & Magnetism
153261
Calculate the magnetic field at the centre of a coil in the form of a square of side 2a carrying a current $I$.
B According to the question, We know that, side of square $=2 \mathrm{a}$ Magnetic field $(B)=\frac{\mu_{0}}{4 \pi \mathrm{r}} \mathrm{I}(\sin \alpha+\sin \beta)$ From the figure, $\alpha= \beta=45^{\circ}$ $\mathrm{B}_{\text {net }} =4 \times \frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{a}}\left[\sin 45+\sin 45^{\circ}\right]$ $\mathrm{B}_{\text {net }} =4 \times \frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{a}}\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right]$ $\mathrm{B}_{\text {net }} =\frac{4 \mu_{0} \mathrm{I}}{4 \pi \mathrm{a}} \frac{2}{\sqrt{2}}$ $\mathrm{~B}_{\text {net }} =\frac{4 \mu_{0} \mathrm{I}}{4 \pi \mathrm{a}} \times \frac{2 \sqrt{2}}{2}$ $\mathrm{~B}_{\text {net }} =\frac{\sqrt{2} \mu_{0} \mathrm{I}}{\pi \mathrm{a}}$
AIIMS-2012
Moving Charges & Magnetism
153262
A steady current is set up in a cubic network composed of wires of equal resistance and length $d$ as shown in figure. What is the magnetic field at the centre $P$ due to the cubic network
C From the given figure, All the diagonally opposite arms will carry same current. So, the net magnetic field at $\mathrm{P}$ will be zero. Because the magnetic field at center $\mathrm{P}$ is cancelled by opposite pairs. Ex. AD \& EF will cancel each other. Therefore, magnetic field at the centre $\mathrm{P}$ due to the cubic network is zero.
