153230
Two long straight wires are set parallel to each other. Each carries a current in the same direction and the separation between them is 2r. The intensity of the magnetic field mid way between them is
1 $\frac{\mu_{0} \mathrm{i}}{\mathrm{r}}$
2 $\frac{4 \mu_{0} \mathrm{i}}{\mathrm{r}}$
3 zero
4 $\frac{\mu_{0} \mathrm{i}}{4 \mathrm{r}}$
Explanation:
C The magnetic field produce at the midpoint $P$, by the current following in the same direction in the two parallel wire will be equal in magnitude but opposite in direction. Therefore, the resultant magnetic field at $\mathrm{P}$ will be zero.
Manipal UGET-2018
Moving Charges & Magnetism
153236
A circular coil carrying current ' $I$ ' has radius ' $R$ ' and magnetic field at the centre is ' $B$ '. At what distance from the centre along the axis of the same coil, the magnetic field will be $\frac{B}{8}$ ?
1 $\mathrm{R} \sqrt{2}$
2 $\mathrm{R} \sqrt{3}$
3 $2 \mathrm{R}$
4 $3 \mathrm{R}$
Explanation:
B We know that magnetic field on the axis of a current carrying coil at distance $\mathrm{x}$ from the center is given by- $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{IR}^{2}}{2\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}$ And magnetic field at the center of a current carrying wire- $\because \quad \mathrm{B} =\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$ $\mathrm{B}_{1}=\mathrm{B} / 8$ $\frac{\mu_{0} \mathrm{IR}^{2}}{2\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}=\frac{\mu_{0} \mathrm{I}}{16 \mathrm{R}}$ $8 \mathrm{R}^{3} =\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{3 / 2}$ $2 \mathrm{R} =\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{1 / 2}$ $4 \mathrm{R}^{2} =\mathrm{R}^{2}+\mathrm{x}^{2}$ $\mathrm{x}^{2} =3 \mathrm{R}^{2}$ $\mathrm{x} =\mathrm{R} \sqrt{3}$
MHT-CET 2017
Moving Charges & Magnetism
153237
Two long parallel wires placed $0.08 \mathrm{~m}$ apart carry currents $3 \mathrm{~A}$ and $5 \mathrm{~A}$ in the same direction. What is the distance from the conductor carrying larger current to the point where the resultant magnetic field is zero?
1 $0.5 \mathrm{~m}$
2 $0.04 \mathrm{~m}$
3 $0.05 \mathrm{~m}$
4 $0.4 \mathrm{~m}$
Explanation:
C Let magnetic field at point $\mathrm{P}$ is zero. Magnetic field at point ' $\mathrm{P}$ ' due to wire 1 $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}_{1}}{2 \pi(0.08-\mathrm{x})} \text { into the plane of paper }$ Magnetic field at point ' $\mathrm{p}$ ' due to wire 2 $B_{2}=\frac{\mu_{0} I_{2}}{2 \pi(x)} \text { out of the plane of paper }$ $\therefore \quad \mathrm{B}_{1}=\mathrm{B}_{2}$ (Because resultant magnetic field zero) $\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \pi(0.08-\mathrm{x})}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{2}}{2 \pi(\mathrm{x})}$ $\frac{\mathrm{I}_{1}}{0.08-\mathrm{x}}=\frac{\mathrm{I}_{2}}{\mathrm{x}}$ \(\frac{3}{0.08-x}=\frac{5}{x}\) Where, \(\mathrm{I}_1=3 \mathrm{~A}\) \(\mathrm{I}_2=5 \mathrm{~A}\) $\mathrm{x}=0.05 \mathrm{~m}$
JCECE-2017
Moving Charges & Magnetism
153238
A 250-turn rectangular coil of length $2.1 \mathrm{~cm}$ and width $1.25 \mathrm{~cm}$ carries a current of $85 \mu \mathrm{A}$ and subjected to a magnetic field of strength $0.85 \mathrm{~T}$. Work done for rotating the coil by $180^{\circ}$ against the torque is
153230
Two long straight wires are set parallel to each other. Each carries a current in the same direction and the separation between them is 2r. The intensity of the magnetic field mid way between them is
1 $\frac{\mu_{0} \mathrm{i}}{\mathrm{r}}$
2 $\frac{4 \mu_{0} \mathrm{i}}{\mathrm{r}}$
3 zero
4 $\frac{\mu_{0} \mathrm{i}}{4 \mathrm{r}}$
Explanation:
C The magnetic field produce at the midpoint $P$, by the current following in the same direction in the two parallel wire will be equal in magnitude but opposite in direction. Therefore, the resultant magnetic field at $\mathrm{P}$ will be zero.
Manipal UGET-2018
Moving Charges & Magnetism
153236
A circular coil carrying current ' $I$ ' has radius ' $R$ ' and magnetic field at the centre is ' $B$ '. At what distance from the centre along the axis of the same coil, the magnetic field will be $\frac{B}{8}$ ?
1 $\mathrm{R} \sqrt{2}$
2 $\mathrm{R} \sqrt{3}$
3 $2 \mathrm{R}$
4 $3 \mathrm{R}$
Explanation:
B We know that magnetic field on the axis of a current carrying coil at distance $\mathrm{x}$ from the center is given by- $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{IR}^{2}}{2\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}$ And magnetic field at the center of a current carrying wire- $\because \quad \mathrm{B} =\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$ $\mathrm{B}_{1}=\mathrm{B} / 8$ $\frac{\mu_{0} \mathrm{IR}^{2}}{2\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}=\frac{\mu_{0} \mathrm{I}}{16 \mathrm{R}}$ $8 \mathrm{R}^{3} =\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{3 / 2}$ $2 \mathrm{R} =\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{1 / 2}$ $4 \mathrm{R}^{2} =\mathrm{R}^{2}+\mathrm{x}^{2}$ $\mathrm{x}^{2} =3 \mathrm{R}^{2}$ $\mathrm{x} =\mathrm{R} \sqrt{3}$
MHT-CET 2017
Moving Charges & Magnetism
153237
Two long parallel wires placed $0.08 \mathrm{~m}$ apart carry currents $3 \mathrm{~A}$ and $5 \mathrm{~A}$ in the same direction. What is the distance from the conductor carrying larger current to the point where the resultant magnetic field is zero?
1 $0.5 \mathrm{~m}$
2 $0.04 \mathrm{~m}$
3 $0.05 \mathrm{~m}$
4 $0.4 \mathrm{~m}$
Explanation:
C Let magnetic field at point $\mathrm{P}$ is zero. Magnetic field at point ' $\mathrm{P}$ ' due to wire 1 $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}_{1}}{2 \pi(0.08-\mathrm{x})} \text { into the plane of paper }$ Magnetic field at point ' $\mathrm{p}$ ' due to wire 2 $B_{2}=\frac{\mu_{0} I_{2}}{2 \pi(x)} \text { out of the plane of paper }$ $\therefore \quad \mathrm{B}_{1}=\mathrm{B}_{2}$ (Because resultant magnetic field zero) $\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \pi(0.08-\mathrm{x})}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{2}}{2 \pi(\mathrm{x})}$ $\frac{\mathrm{I}_{1}}{0.08-\mathrm{x}}=\frac{\mathrm{I}_{2}}{\mathrm{x}}$ \(\frac{3}{0.08-x}=\frac{5}{x}\) Where, \(\mathrm{I}_1=3 \mathrm{~A}\) \(\mathrm{I}_2=5 \mathrm{~A}\) $\mathrm{x}=0.05 \mathrm{~m}$
JCECE-2017
Moving Charges & Magnetism
153238
A 250-turn rectangular coil of length $2.1 \mathrm{~cm}$ and width $1.25 \mathrm{~cm}$ carries a current of $85 \mu \mathrm{A}$ and subjected to a magnetic field of strength $0.85 \mathrm{~T}$. Work done for rotating the coil by $180^{\circ}$ against the torque is
153230
Two long straight wires are set parallel to each other. Each carries a current in the same direction and the separation between them is 2r. The intensity of the magnetic field mid way between them is
1 $\frac{\mu_{0} \mathrm{i}}{\mathrm{r}}$
2 $\frac{4 \mu_{0} \mathrm{i}}{\mathrm{r}}$
3 zero
4 $\frac{\mu_{0} \mathrm{i}}{4 \mathrm{r}}$
Explanation:
C The magnetic field produce at the midpoint $P$, by the current following in the same direction in the two parallel wire will be equal in magnitude but opposite in direction. Therefore, the resultant magnetic field at $\mathrm{P}$ will be zero.
Manipal UGET-2018
Moving Charges & Magnetism
153236
A circular coil carrying current ' $I$ ' has radius ' $R$ ' and magnetic field at the centre is ' $B$ '. At what distance from the centre along the axis of the same coil, the magnetic field will be $\frac{B}{8}$ ?
1 $\mathrm{R} \sqrt{2}$
2 $\mathrm{R} \sqrt{3}$
3 $2 \mathrm{R}$
4 $3 \mathrm{R}$
Explanation:
B We know that magnetic field on the axis of a current carrying coil at distance $\mathrm{x}$ from the center is given by- $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{IR}^{2}}{2\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}$ And magnetic field at the center of a current carrying wire- $\because \quad \mathrm{B} =\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$ $\mathrm{B}_{1}=\mathrm{B} / 8$ $\frac{\mu_{0} \mathrm{IR}^{2}}{2\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}=\frac{\mu_{0} \mathrm{I}}{16 \mathrm{R}}$ $8 \mathrm{R}^{3} =\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{3 / 2}$ $2 \mathrm{R} =\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{1 / 2}$ $4 \mathrm{R}^{2} =\mathrm{R}^{2}+\mathrm{x}^{2}$ $\mathrm{x}^{2} =3 \mathrm{R}^{2}$ $\mathrm{x} =\mathrm{R} \sqrt{3}$
MHT-CET 2017
Moving Charges & Magnetism
153237
Two long parallel wires placed $0.08 \mathrm{~m}$ apart carry currents $3 \mathrm{~A}$ and $5 \mathrm{~A}$ in the same direction. What is the distance from the conductor carrying larger current to the point where the resultant magnetic field is zero?
1 $0.5 \mathrm{~m}$
2 $0.04 \mathrm{~m}$
3 $0.05 \mathrm{~m}$
4 $0.4 \mathrm{~m}$
Explanation:
C Let magnetic field at point $\mathrm{P}$ is zero. Magnetic field at point ' $\mathrm{P}$ ' due to wire 1 $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}_{1}}{2 \pi(0.08-\mathrm{x})} \text { into the plane of paper }$ Magnetic field at point ' $\mathrm{p}$ ' due to wire 2 $B_{2}=\frac{\mu_{0} I_{2}}{2 \pi(x)} \text { out of the plane of paper }$ $\therefore \quad \mathrm{B}_{1}=\mathrm{B}_{2}$ (Because resultant magnetic field zero) $\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \pi(0.08-\mathrm{x})}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{2}}{2 \pi(\mathrm{x})}$ $\frac{\mathrm{I}_{1}}{0.08-\mathrm{x}}=\frac{\mathrm{I}_{2}}{\mathrm{x}}$ \(\frac{3}{0.08-x}=\frac{5}{x}\) Where, \(\mathrm{I}_1=3 \mathrm{~A}\) \(\mathrm{I}_2=5 \mathrm{~A}\) $\mathrm{x}=0.05 \mathrm{~m}$
JCECE-2017
Moving Charges & Magnetism
153238
A 250-turn rectangular coil of length $2.1 \mathrm{~cm}$ and width $1.25 \mathrm{~cm}$ carries a current of $85 \mu \mathrm{A}$ and subjected to a magnetic field of strength $0.85 \mathrm{~T}$. Work done for rotating the coil by $180^{\circ}$ against the torque is
153230
Two long straight wires are set parallel to each other. Each carries a current in the same direction and the separation between them is 2r. The intensity of the magnetic field mid way between them is
1 $\frac{\mu_{0} \mathrm{i}}{\mathrm{r}}$
2 $\frac{4 \mu_{0} \mathrm{i}}{\mathrm{r}}$
3 zero
4 $\frac{\mu_{0} \mathrm{i}}{4 \mathrm{r}}$
Explanation:
C The magnetic field produce at the midpoint $P$, by the current following in the same direction in the two parallel wire will be equal in magnitude but opposite in direction. Therefore, the resultant magnetic field at $\mathrm{P}$ will be zero.
Manipal UGET-2018
Moving Charges & Magnetism
153236
A circular coil carrying current ' $I$ ' has radius ' $R$ ' and magnetic field at the centre is ' $B$ '. At what distance from the centre along the axis of the same coil, the magnetic field will be $\frac{B}{8}$ ?
1 $\mathrm{R} \sqrt{2}$
2 $\mathrm{R} \sqrt{3}$
3 $2 \mathrm{R}$
4 $3 \mathrm{R}$
Explanation:
B We know that magnetic field on the axis of a current carrying coil at distance $\mathrm{x}$ from the center is given by- $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{IR}^{2}}{2\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}$ And magnetic field at the center of a current carrying wire- $\because \quad \mathrm{B} =\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$ $\mathrm{B}_{1}=\mathrm{B} / 8$ $\frac{\mu_{0} \mathrm{IR}^{2}}{2\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}=\frac{\mu_{0} \mathrm{I}}{16 \mathrm{R}}$ $8 \mathrm{R}^{3} =\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{3 / 2}$ $2 \mathrm{R} =\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{1 / 2}$ $4 \mathrm{R}^{2} =\mathrm{R}^{2}+\mathrm{x}^{2}$ $\mathrm{x}^{2} =3 \mathrm{R}^{2}$ $\mathrm{x} =\mathrm{R} \sqrt{3}$
MHT-CET 2017
Moving Charges & Magnetism
153237
Two long parallel wires placed $0.08 \mathrm{~m}$ apart carry currents $3 \mathrm{~A}$ and $5 \mathrm{~A}$ in the same direction. What is the distance from the conductor carrying larger current to the point where the resultant magnetic field is zero?
1 $0.5 \mathrm{~m}$
2 $0.04 \mathrm{~m}$
3 $0.05 \mathrm{~m}$
4 $0.4 \mathrm{~m}$
Explanation:
C Let magnetic field at point $\mathrm{P}$ is zero. Magnetic field at point ' $\mathrm{P}$ ' due to wire 1 $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}_{1}}{2 \pi(0.08-\mathrm{x})} \text { into the plane of paper }$ Magnetic field at point ' $\mathrm{p}$ ' due to wire 2 $B_{2}=\frac{\mu_{0} I_{2}}{2 \pi(x)} \text { out of the plane of paper }$ $\therefore \quad \mathrm{B}_{1}=\mathrm{B}_{2}$ (Because resultant magnetic field zero) $\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \pi(0.08-\mathrm{x})}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{2}}{2 \pi(\mathrm{x})}$ $\frac{\mathrm{I}_{1}}{0.08-\mathrm{x}}=\frac{\mathrm{I}_{2}}{\mathrm{x}}$ \(\frac{3}{0.08-x}=\frac{5}{x}\) Where, \(\mathrm{I}_1=3 \mathrm{~A}\) \(\mathrm{I}_2=5 \mathrm{~A}\) $\mathrm{x}=0.05 \mathrm{~m}$
JCECE-2017
Moving Charges & Magnetism
153238
A 250-turn rectangular coil of length $2.1 \mathrm{~cm}$ and width $1.25 \mathrm{~cm}$ carries a current of $85 \mu \mathrm{A}$ and subjected to a magnetic field of strength $0.85 \mathrm{~T}$. Work done for rotating the coil by $180^{\circ}$ against the torque is
