362651
A wire carrying current \(i\) and other parallel wire carrying \(2i\) in the same direction produce a magnetic field \(B\) at the mid point. What will be the field when \(2i\) current is switched off?
362652
The current in straight wire if the magnetic field \({10^{ - 6}}Wb{m^{ - 2}}\) produced at 0.02\(m\) away from it is
1 0.1\(A\)
2 1\(A\)
3 zero
4 10\(A\)
Explanation:
In case of straight conductor of infinite length, the magnetic field is given by \(B = \frac{{{\mu _0}}}{{4\pi }} \times \frac{{2i}}{a}\) \( \Rightarrow {10^{ - 6}} = \frac{{{{10}^{ - 7}} \times 2 \times i}}{{0.02}}\) \( \Rightarrow i = 0.1A\)
PHXII04:MOVING CHARGES AND MAGNETISM
362653
A horizontal overhead powerline is at height of \(4\;m\) from the ground and carries a current of \(100\,A\) from east to west. The magnetic field directly below it on the ground is (\({\mu _0} = 4\pi \times {10^{ - 7}}T\,m{A^{ - 1}}\))
1 \(2.5 \times {10^{ - 7}}\;T\) southward
2 \(5 \times {10^{ - 6}}\;T\) northward
3 \(5 \times {10^{ - 6}}\;T\) southward
4 \(2.5 \times {10^{ - 7}}\;T\) northward
Explanation:
The magnetic field is \(B = \frac{{{\mu _0}}}{{4\pi }}\frac{{2I}}{r} = {10^{ - 7}} \times \frac{{2 \times 100}}{4} = 5 \times {10^{ - 6}}\;T\)
PHXII04:MOVING CHARGES AND MAGNETISM
362654
A long straight wire along the \(z\)-axis carries a current \(i\) in the negative \(z\)-direction. The magnetic vector field \(B\) at a point having coordinates \((x,y)\) in the \(z = 0\) plane is
The magnetic field vector is shown in the figure. \(\begin{aligned}& \bar{B}=\dfrac{\mu_{o} i}{2 \pi r} \hat{n} \\& \hat{n}=-|\hat{n}| \cos \theta \hat{j}+|\hat{n}| \sin \theta \hat{i} \\& \hat{n}=-\dfrac{x}{r} \hat{j}+\dfrac{y}{r} \hat{i} \\& \bar{B}=\dfrac{\mu_{o} i(y \hat{i}-x \hat{j})}{2 \pi\left(x^{2}+y^{2}\right)}\end{aligned}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362655
The position of point \({P}\) from wire \({B}\) where net magnetic field is zero, due to following current distribution, will be
1 \(5\,cm\) at a point lying outside the wires
2 \(3\,cm\) at a point lying between the wires
3 \(1\,cm\) at a point lying between the wires
4 \(2\,cm\) at a point lying outside the wires
Explanation:
For \({B}\) to be zero at \({P}\), \(B_{1}=B_{2}\) \({\dfrac{\mu_{0}(12.5)}{2 \pi(6-x)}=\dfrac{\mu_{0}(2.5)}{2 \pi x}}\) \({\dfrac{5}{1}=\dfrac{6-x}{x}}\) \({5 x=6-x}\) \({x=1 {~cm}}\) Thus \({B}\) will be zero at point \({P}\) at a distance \({1 cm}\) from wire carrying \({2.5 A}\) current.
362651
A wire carrying current \(i\) and other parallel wire carrying \(2i\) in the same direction produce a magnetic field \(B\) at the mid point. What will be the field when \(2i\) current is switched off?
362652
The current in straight wire if the magnetic field \({10^{ - 6}}Wb{m^{ - 2}}\) produced at 0.02\(m\) away from it is
1 0.1\(A\)
2 1\(A\)
3 zero
4 10\(A\)
Explanation:
In case of straight conductor of infinite length, the magnetic field is given by \(B = \frac{{{\mu _0}}}{{4\pi }} \times \frac{{2i}}{a}\) \( \Rightarrow {10^{ - 6}} = \frac{{{{10}^{ - 7}} \times 2 \times i}}{{0.02}}\) \( \Rightarrow i = 0.1A\)
PHXII04:MOVING CHARGES AND MAGNETISM
362653
A horizontal overhead powerline is at height of \(4\;m\) from the ground and carries a current of \(100\,A\) from east to west. The magnetic field directly below it on the ground is (\({\mu _0} = 4\pi \times {10^{ - 7}}T\,m{A^{ - 1}}\))
1 \(2.5 \times {10^{ - 7}}\;T\) southward
2 \(5 \times {10^{ - 6}}\;T\) northward
3 \(5 \times {10^{ - 6}}\;T\) southward
4 \(2.5 \times {10^{ - 7}}\;T\) northward
Explanation:
The magnetic field is \(B = \frac{{{\mu _0}}}{{4\pi }}\frac{{2I}}{r} = {10^{ - 7}} \times \frac{{2 \times 100}}{4} = 5 \times {10^{ - 6}}\;T\)
PHXII04:MOVING CHARGES AND MAGNETISM
362654
A long straight wire along the \(z\)-axis carries a current \(i\) in the negative \(z\)-direction. The magnetic vector field \(B\) at a point having coordinates \((x,y)\) in the \(z = 0\) plane is
The magnetic field vector is shown in the figure. \(\begin{aligned}& \bar{B}=\dfrac{\mu_{o} i}{2 \pi r} \hat{n} \\& \hat{n}=-|\hat{n}| \cos \theta \hat{j}+|\hat{n}| \sin \theta \hat{i} \\& \hat{n}=-\dfrac{x}{r} \hat{j}+\dfrac{y}{r} \hat{i} \\& \bar{B}=\dfrac{\mu_{o} i(y \hat{i}-x \hat{j})}{2 \pi\left(x^{2}+y^{2}\right)}\end{aligned}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362655
The position of point \({P}\) from wire \({B}\) where net magnetic field is zero, due to following current distribution, will be
1 \(5\,cm\) at a point lying outside the wires
2 \(3\,cm\) at a point lying between the wires
3 \(1\,cm\) at a point lying between the wires
4 \(2\,cm\) at a point lying outside the wires
Explanation:
For \({B}\) to be zero at \({P}\), \(B_{1}=B_{2}\) \({\dfrac{\mu_{0}(12.5)}{2 \pi(6-x)}=\dfrac{\mu_{0}(2.5)}{2 \pi x}}\) \({\dfrac{5}{1}=\dfrac{6-x}{x}}\) \({5 x=6-x}\) \({x=1 {~cm}}\) Thus \({B}\) will be zero at point \({P}\) at a distance \({1 cm}\) from wire carrying \({2.5 A}\) current.
362651
A wire carrying current \(i\) and other parallel wire carrying \(2i\) in the same direction produce a magnetic field \(B\) at the mid point. What will be the field when \(2i\) current is switched off?
362652
The current in straight wire if the magnetic field \({10^{ - 6}}Wb{m^{ - 2}}\) produced at 0.02\(m\) away from it is
1 0.1\(A\)
2 1\(A\)
3 zero
4 10\(A\)
Explanation:
In case of straight conductor of infinite length, the magnetic field is given by \(B = \frac{{{\mu _0}}}{{4\pi }} \times \frac{{2i}}{a}\) \( \Rightarrow {10^{ - 6}} = \frac{{{{10}^{ - 7}} \times 2 \times i}}{{0.02}}\) \( \Rightarrow i = 0.1A\)
PHXII04:MOVING CHARGES AND MAGNETISM
362653
A horizontal overhead powerline is at height of \(4\;m\) from the ground and carries a current of \(100\,A\) from east to west. The magnetic field directly below it on the ground is (\({\mu _0} = 4\pi \times {10^{ - 7}}T\,m{A^{ - 1}}\))
1 \(2.5 \times {10^{ - 7}}\;T\) southward
2 \(5 \times {10^{ - 6}}\;T\) northward
3 \(5 \times {10^{ - 6}}\;T\) southward
4 \(2.5 \times {10^{ - 7}}\;T\) northward
Explanation:
The magnetic field is \(B = \frac{{{\mu _0}}}{{4\pi }}\frac{{2I}}{r} = {10^{ - 7}} \times \frac{{2 \times 100}}{4} = 5 \times {10^{ - 6}}\;T\)
PHXII04:MOVING CHARGES AND MAGNETISM
362654
A long straight wire along the \(z\)-axis carries a current \(i\) in the negative \(z\)-direction. The magnetic vector field \(B\) at a point having coordinates \((x,y)\) in the \(z = 0\) plane is
The magnetic field vector is shown in the figure. \(\begin{aligned}& \bar{B}=\dfrac{\mu_{o} i}{2 \pi r} \hat{n} \\& \hat{n}=-|\hat{n}| \cos \theta \hat{j}+|\hat{n}| \sin \theta \hat{i} \\& \hat{n}=-\dfrac{x}{r} \hat{j}+\dfrac{y}{r} \hat{i} \\& \bar{B}=\dfrac{\mu_{o} i(y \hat{i}-x \hat{j})}{2 \pi\left(x^{2}+y^{2}\right)}\end{aligned}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362655
The position of point \({P}\) from wire \({B}\) where net magnetic field is zero, due to following current distribution, will be
1 \(5\,cm\) at a point lying outside the wires
2 \(3\,cm\) at a point lying between the wires
3 \(1\,cm\) at a point lying between the wires
4 \(2\,cm\) at a point lying outside the wires
Explanation:
For \({B}\) to be zero at \({P}\), \(B_{1}=B_{2}\) \({\dfrac{\mu_{0}(12.5)}{2 \pi(6-x)}=\dfrac{\mu_{0}(2.5)}{2 \pi x}}\) \({\dfrac{5}{1}=\dfrac{6-x}{x}}\) \({5 x=6-x}\) \({x=1 {~cm}}\) Thus \({B}\) will be zero at point \({P}\) at a distance \({1 cm}\) from wire carrying \({2.5 A}\) current.
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PHXII04:MOVING CHARGES AND MAGNETISM
362651
A wire carrying current \(i\) and other parallel wire carrying \(2i\) in the same direction produce a magnetic field \(B\) at the mid point. What will be the field when \(2i\) current is switched off?
362652
The current in straight wire if the magnetic field \({10^{ - 6}}Wb{m^{ - 2}}\) produced at 0.02\(m\) away from it is
1 0.1\(A\)
2 1\(A\)
3 zero
4 10\(A\)
Explanation:
In case of straight conductor of infinite length, the magnetic field is given by \(B = \frac{{{\mu _0}}}{{4\pi }} \times \frac{{2i}}{a}\) \( \Rightarrow {10^{ - 6}} = \frac{{{{10}^{ - 7}} \times 2 \times i}}{{0.02}}\) \( \Rightarrow i = 0.1A\)
PHXII04:MOVING CHARGES AND MAGNETISM
362653
A horizontal overhead powerline is at height of \(4\;m\) from the ground and carries a current of \(100\,A\) from east to west. The magnetic field directly below it on the ground is (\({\mu _0} = 4\pi \times {10^{ - 7}}T\,m{A^{ - 1}}\))
1 \(2.5 \times {10^{ - 7}}\;T\) southward
2 \(5 \times {10^{ - 6}}\;T\) northward
3 \(5 \times {10^{ - 6}}\;T\) southward
4 \(2.5 \times {10^{ - 7}}\;T\) northward
Explanation:
The magnetic field is \(B = \frac{{{\mu _0}}}{{4\pi }}\frac{{2I}}{r} = {10^{ - 7}} \times \frac{{2 \times 100}}{4} = 5 \times {10^{ - 6}}\;T\)
PHXII04:MOVING CHARGES AND MAGNETISM
362654
A long straight wire along the \(z\)-axis carries a current \(i\) in the negative \(z\)-direction. The magnetic vector field \(B\) at a point having coordinates \((x,y)\) in the \(z = 0\) plane is
The magnetic field vector is shown in the figure. \(\begin{aligned}& \bar{B}=\dfrac{\mu_{o} i}{2 \pi r} \hat{n} \\& \hat{n}=-|\hat{n}| \cos \theta \hat{j}+|\hat{n}| \sin \theta \hat{i} \\& \hat{n}=-\dfrac{x}{r} \hat{j}+\dfrac{y}{r} \hat{i} \\& \bar{B}=\dfrac{\mu_{o} i(y \hat{i}-x \hat{j})}{2 \pi\left(x^{2}+y^{2}\right)}\end{aligned}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362655
The position of point \({P}\) from wire \({B}\) where net magnetic field is zero, due to following current distribution, will be
1 \(5\,cm\) at a point lying outside the wires
2 \(3\,cm\) at a point lying between the wires
3 \(1\,cm\) at a point lying between the wires
4 \(2\,cm\) at a point lying outside the wires
Explanation:
For \({B}\) to be zero at \({P}\), \(B_{1}=B_{2}\) \({\dfrac{\mu_{0}(12.5)}{2 \pi(6-x)}=\dfrac{\mu_{0}(2.5)}{2 \pi x}}\) \({\dfrac{5}{1}=\dfrac{6-x}{x}}\) \({5 x=6-x}\) \({x=1 {~cm}}\) Thus \({B}\) will be zero at point \({P}\) at a distance \({1 cm}\) from wire carrying \({2.5 A}\) current.
362651
A wire carrying current \(i\) and other parallel wire carrying \(2i\) in the same direction produce a magnetic field \(B\) at the mid point. What will be the field when \(2i\) current is switched off?
362652
The current in straight wire if the magnetic field \({10^{ - 6}}Wb{m^{ - 2}}\) produced at 0.02\(m\) away from it is
1 0.1\(A\)
2 1\(A\)
3 zero
4 10\(A\)
Explanation:
In case of straight conductor of infinite length, the magnetic field is given by \(B = \frac{{{\mu _0}}}{{4\pi }} \times \frac{{2i}}{a}\) \( \Rightarrow {10^{ - 6}} = \frac{{{{10}^{ - 7}} \times 2 \times i}}{{0.02}}\) \( \Rightarrow i = 0.1A\)
PHXII04:MOVING CHARGES AND MAGNETISM
362653
A horizontal overhead powerline is at height of \(4\;m\) from the ground and carries a current of \(100\,A\) from east to west. The magnetic field directly below it on the ground is (\({\mu _0} = 4\pi \times {10^{ - 7}}T\,m{A^{ - 1}}\))
1 \(2.5 \times {10^{ - 7}}\;T\) southward
2 \(5 \times {10^{ - 6}}\;T\) northward
3 \(5 \times {10^{ - 6}}\;T\) southward
4 \(2.5 \times {10^{ - 7}}\;T\) northward
Explanation:
The magnetic field is \(B = \frac{{{\mu _0}}}{{4\pi }}\frac{{2I}}{r} = {10^{ - 7}} \times \frac{{2 \times 100}}{4} = 5 \times {10^{ - 6}}\;T\)
PHXII04:MOVING CHARGES AND MAGNETISM
362654
A long straight wire along the \(z\)-axis carries a current \(i\) in the negative \(z\)-direction. The magnetic vector field \(B\) at a point having coordinates \((x,y)\) in the \(z = 0\) plane is
The magnetic field vector is shown in the figure. \(\begin{aligned}& \bar{B}=\dfrac{\mu_{o} i}{2 \pi r} \hat{n} \\& \hat{n}=-|\hat{n}| \cos \theta \hat{j}+|\hat{n}| \sin \theta \hat{i} \\& \hat{n}=-\dfrac{x}{r} \hat{j}+\dfrac{y}{r} \hat{i} \\& \bar{B}=\dfrac{\mu_{o} i(y \hat{i}-x \hat{j})}{2 \pi\left(x^{2}+y^{2}\right)}\end{aligned}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362655
The position of point \({P}\) from wire \({B}\) where net magnetic field is zero, due to following current distribution, will be
1 \(5\,cm\) at a point lying outside the wires
2 \(3\,cm\) at a point lying between the wires
3 \(1\,cm\) at a point lying between the wires
4 \(2\,cm\) at a point lying outside the wires
Explanation:
For \({B}\) to be zero at \({P}\), \(B_{1}=B_{2}\) \({\dfrac{\mu_{0}(12.5)}{2 \pi(6-x)}=\dfrac{\mu_{0}(2.5)}{2 \pi x}}\) \({\dfrac{5}{1}=\dfrac{6-x}{x}}\) \({5 x=6-x}\) \({x=1 {~cm}}\) Thus \({B}\) will be zero at point \({P}\) at a distance \({1 cm}\) from wire carrying \({2.5 A}\) current.
