153386
Electron moves at right angles to a magnetic field of $1.5 \times 10^{-2}$ tesla with speed of $6 \times 10^{7} \mathrm{~m} / \mathrm{s}$. If the specific charge of the electron is $1.7 \times$ $10^{11} \mathrm{C} / \mathrm{kg}$. The radius of circular path will be
153389
A straight wire of mass $300 \mathrm{~g}$ and length $2.5 \mathrm{~m}$ carries a current of $3.5 \mathrm{~A}$. It is suspended in mid-air by a uniform horizontal magnetic field $B$. What is the magnitude of the magnetic field?
1 $0.654 \mathrm{~T}$
2 $0.336 \mathrm{~T}$
3 $1.576 \mathrm{~T}$
4 $0.939 \mathrm{~T}$
Explanation:
B Here weight of the wire is balanced by the extreme magnetic force- So, $\quad$ B il $=\mathrm{mg}$ $\mathrm{B}=\frac{\mathrm{mg}}{\mathrm{i} l}=\frac{300 \times 10^{-3} \times 10}{3.5 \times 2.5}$ $\mathrm{~B}=0.34 \mathrm{~T}$
J and K CET-2014
Moving Charges & Magnetism
153399
Magnetic field due to $0.1 \mathrm{~A}$ current flowing through a circular coil of radius $0.1 \mathrm{~m}$ and 1000 turns at the centre of the coil is
153401
Two wires are held perpendicular to the plane of paper and are $5 \mathrm{~m}$ apart. They carry currents of $2.5 \mathrm{~A}$ and $5 \mathrm{~A}$ in same direction. Then, the magnetic field strength $(B)$ at a point midway between the wires will be
153386
Electron moves at right angles to a magnetic field of $1.5 \times 10^{-2}$ tesla with speed of $6 \times 10^{7} \mathrm{~m} / \mathrm{s}$. If the specific charge of the electron is $1.7 \times$ $10^{11} \mathrm{C} / \mathrm{kg}$. The radius of circular path will be
153389
A straight wire of mass $300 \mathrm{~g}$ and length $2.5 \mathrm{~m}$ carries a current of $3.5 \mathrm{~A}$. It is suspended in mid-air by a uniform horizontal magnetic field $B$. What is the magnitude of the magnetic field?
1 $0.654 \mathrm{~T}$
2 $0.336 \mathrm{~T}$
3 $1.576 \mathrm{~T}$
4 $0.939 \mathrm{~T}$
Explanation:
B Here weight of the wire is balanced by the extreme magnetic force- So, $\quad$ B il $=\mathrm{mg}$ $\mathrm{B}=\frac{\mathrm{mg}}{\mathrm{i} l}=\frac{300 \times 10^{-3} \times 10}{3.5 \times 2.5}$ $\mathrm{~B}=0.34 \mathrm{~T}$
J and K CET-2014
Moving Charges & Magnetism
153399
Magnetic field due to $0.1 \mathrm{~A}$ current flowing through a circular coil of radius $0.1 \mathrm{~m}$ and 1000 turns at the centre of the coil is
153401
Two wires are held perpendicular to the plane of paper and are $5 \mathrm{~m}$ apart. They carry currents of $2.5 \mathrm{~A}$ and $5 \mathrm{~A}$ in same direction. Then, the magnetic field strength $(B)$ at a point midway between the wires will be
153386
Electron moves at right angles to a magnetic field of $1.5 \times 10^{-2}$ tesla with speed of $6 \times 10^{7} \mathrm{~m} / \mathrm{s}$. If the specific charge of the electron is $1.7 \times$ $10^{11} \mathrm{C} / \mathrm{kg}$. The radius of circular path will be
153389
A straight wire of mass $300 \mathrm{~g}$ and length $2.5 \mathrm{~m}$ carries a current of $3.5 \mathrm{~A}$. It is suspended in mid-air by a uniform horizontal magnetic field $B$. What is the magnitude of the magnetic field?
1 $0.654 \mathrm{~T}$
2 $0.336 \mathrm{~T}$
3 $1.576 \mathrm{~T}$
4 $0.939 \mathrm{~T}$
Explanation:
B Here weight of the wire is balanced by the extreme magnetic force- So, $\quad$ B il $=\mathrm{mg}$ $\mathrm{B}=\frac{\mathrm{mg}}{\mathrm{i} l}=\frac{300 \times 10^{-3} \times 10}{3.5 \times 2.5}$ $\mathrm{~B}=0.34 \mathrm{~T}$
J and K CET-2014
Moving Charges & Magnetism
153399
Magnetic field due to $0.1 \mathrm{~A}$ current flowing through a circular coil of radius $0.1 \mathrm{~m}$ and 1000 turns at the centre of the coil is
153401
Two wires are held perpendicular to the plane of paper and are $5 \mathrm{~m}$ apart. They carry currents of $2.5 \mathrm{~A}$ and $5 \mathrm{~A}$ in same direction. Then, the magnetic field strength $(B)$ at a point midway between the wires will be
153386
Electron moves at right angles to a magnetic field of $1.5 \times 10^{-2}$ tesla with speed of $6 \times 10^{7} \mathrm{~m} / \mathrm{s}$. If the specific charge of the electron is $1.7 \times$ $10^{11} \mathrm{C} / \mathrm{kg}$. The radius of circular path will be
153389
A straight wire of mass $300 \mathrm{~g}$ and length $2.5 \mathrm{~m}$ carries a current of $3.5 \mathrm{~A}$. It is suspended in mid-air by a uniform horizontal magnetic field $B$. What is the magnitude of the magnetic field?
1 $0.654 \mathrm{~T}$
2 $0.336 \mathrm{~T}$
3 $1.576 \mathrm{~T}$
4 $0.939 \mathrm{~T}$
Explanation:
B Here weight of the wire is balanced by the extreme magnetic force- So, $\quad$ B il $=\mathrm{mg}$ $\mathrm{B}=\frac{\mathrm{mg}}{\mathrm{i} l}=\frac{300 \times 10^{-3} \times 10}{3.5 \times 2.5}$ $\mathrm{~B}=0.34 \mathrm{~T}$
J and K CET-2014
Moving Charges & Magnetism
153399
Magnetic field due to $0.1 \mathrm{~A}$ current flowing through a circular coil of radius $0.1 \mathrm{~m}$ and 1000 turns at the centre of the coil is
153401
Two wires are held perpendicular to the plane of paper and are $5 \mathrm{~m}$ apart. They carry currents of $2.5 \mathrm{~A}$ and $5 \mathrm{~A}$ in same direction. Then, the magnetic field strength $(B)$ at a point midway between the wires will be