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Moving Charges & Magnetism

153370 When a positively charged particle enters a uniform magnetic field with uniform velocity, its trajectory can be
1.a straight line 2. a circle 3 . a helix

1 (1) only

2 (1) or (2)

3 (1) or (3)

4 Any one of (1), (2) and (3)

Explanation:

D As we know that, magnetic force on a moving charge in a magnetic field is,
$F = qvB$
- When the particle moves along the direction of magnetic field then,
$F = 0$
Thus, the particle moves in a straight line along the incident direction.
- If the particle moves perpendicular to the direction of magnetic field,
Thus, the trajectory of the particle is circular.
- If the direction of velocity has both parallel and perpendicular component to the direction of magnetic field, the parallel component tends to move along the direction of magnetic field. While the perpendicular component moves the particle in a circle, thus the trajectory of the particle is helix.

EAMCET-2006

Moving Charges & Magnetism

153378 A square coil of side a carries a current $I$ . The magnetic field at the centre of the coil is

1

\frac{μ_{0} I}{a π}

2

\frac{\sqrt{2} μ_{0} I}{a π}

3

\frac{μ_{0} I}{\sqrt{2} a π}

4

\frac{2 \sqrt{2} μ_{0} I}{a π}

Explanation:

D $B_{total} = 4 B_{side}$
The magnetic field at the centre of the coil due to each side is in the same direction and are equal to each other,
$B_{total} = 4 \times \frac{μ_{o} I}{4 π (\frac{a}{2})} [\sin \frac{π}{4} + \sin \frac{π}{4}]$
$B_{total} = \frac{2 \sqrt{2} μ_{o} I}{a π}$

AIIMS-2012

Moving Charges & Magnetism

153379 Circular loop of a wire and a long straight wire carry currents $I_{c}$ and $I_{e}$ , respectively as shown in figure. Assuming that these are placed in the same plane, the magnetic fields will be zero at the centre of the loop when the separation $H$ is:

1

\frac{I_{e} R}{I_{c} π}

2

\frac{I_{c} R}{I_{e} π}

3

\frac{π I_{c}}{I_{e} R}

4

\frac{I_{e} π}{I_{c} R}

Explanation:

A Magnetic field due to straight wire $= \frac{μ_{o} I_{e}}{2 π H}$ Magnetic field due to circular wire $= \frac{μ_{o} I_{c}}{2 R}$
Now, $\frac{μ_{o} I_{e}}{2 π H} = \frac{μ_{o} I_{c}}{2 R}$
$H = \frac{{RI}_{e}}{π I_{c}}$

AIIMS-2006

Moving Charges & Magnetism

153381 Two concentric coils each of radius equal to $2 π$ cm are placed at right angles to each other. 3 ampere and 4 ampere are the currents flowing in each coil respectively. The magnetic induction in Weber $/ m^{2}$ at the centre of the coils will be.
$(μ_{0} = 4 π \times 10^{- 7} Wb / A . m)$

1

10^{- 5}

2

12 \times 10^{- 5}

3

7 \times 10^{- 5}

4

5 \times 10^{- 5}

Explanation:

D Magnetic field due to circular coil,
$I_{1} = 3 A, I_{2} = 4 A$
Where $r = 2 π cm = 2 π \times 10^{- 2} m$
$B_{1} = \frac{μ_{o} i_{1}}{2 r} = \frac{μ_{o} i_{1}}{2 (2 π \times 10^{- 2})} = \frac{μ_{o} \times 3 \times 10^{2}}{4 π}$
$B_{2} = \frac{μ_{o} i_{2}}{2 (2 π \times 10^{- 2})} = \frac{μ_{o} \times 4 \times 10^{2}}{4 π}$
$B = \sqrt{B_{1}^{2} + B_{2}^{2}} = \frac{μ_{0}}{4 π} \times 5 \times 10^{2} = 10^{- 7} \times 5 \times 10^{2}$
$B = 5 \times 10^{- 5}$

AIIMS-2008

Moving Charges & Magnetism

153370 When a positively charged particle enters a uniform magnetic field with uniform velocity, its trajectory can be
1.a straight line 2. a circle 3 . a helix

1 (1) only

2 (1) or (2)

3 (1) or (3)

4 Any one of (1), (2) and (3)

Explanation:

D As we know that, magnetic force on a moving charge in a magnetic field is,
$F = qvB$
- When the particle moves along the direction of magnetic field then,
$F = 0$
Thus, the particle moves in a straight line along the incident direction.
- If the particle moves perpendicular to the direction of magnetic field,
Thus, the trajectory of the particle is circular.
- If the direction of velocity has both parallel and perpendicular component to the direction of magnetic field, the parallel component tends to move along the direction of magnetic field. While the perpendicular component moves the particle in a circle, thus the trajectory of the particle is helix.

EAMCET-2006

Moving Charges & Magnetism

153378 A square coil of side a carries a current $I$ . The magnetic field at the centre of the coil is

1

\frac{μ_{0} I}{a π}

2

\frac{\sqrt{2} μ_{0} I}{a π}

3

\frac{μ_{0} I}{\sqrt{2} a π}

4

\frac{2 \sqrt{2} μ_{0} I}{a π}

Explanation:

D $B_{total} = 4 B_{side}$
The magnetic field at the centre of the coil due to each side is in the same direction and are equal to each other,
$B_{total} = 4 \times \frac{μ_{o} I}{4 π (\frac{a}{2})} [\sin \frac{π}{4} + \sin \frac{π}{4}]$
$B_{total} = \frac{2 \sqrt{2} μ_{o} I}{a π}$

AIIMS-2012

Moving Charges & Magnetism

153379 Circular loop of a wire and a long straight wire carry currents $I_{c}$ and $I_{e}$ , respectively as shown in figure. Assuming that these are placed in the same plane, the magnetic fields will be zero at the centre of the loop when the separation $H$ is:

1

\frac{I_{e} R}{I_{c} π}

2

\frac{I_{c} R}{I_{e} π}

3

\frac{π I_{c}}{I_{e} R}

4

\frac{I_{e} π}{I_{c} R}

Explanation:

A Magnetic field due to straight wire $= \frac{μ_{o} I_{e}}{2 π H}$ Magnetic field due to circular wire $= \frac{μ_{o} I_{c}}{2 R}$
Now, $\frac{μ_{o} I_{e}}{2 π H} = \frac{μ_{o} I_{c}}{2 R}$
$H = \frac{{RI}_{e}}{π I_{c}}$

AIIMS-2006

Moving Charges & Magnetism

153381 Two concentric coils each of radius equal to $2 π$ cm are placed at right angles to each other. 3 ampere and 4 ampere are the currents flowing in each coil respectively. The magnetic induction in Weber $/ m^{2}$ at the centre of the coils will be.
$(μ_{0} = 4 π \times 10^{- 7} Wb / A . m)$

1

10^{- 5}

2

12 \times 10^{- 5}

3

7 \times 10^{- 5}

4

5 \times 10^{- 5}

Explanation:

D Magnetic field due to circular coil,
$I_{1} = 3 A, I_{2} = 4 A$
Where $r = 2 π cm = 2 π \times 10^{- 2} m$
$B_{1} = \frac{μ_{o} i_{1}}{2 r} = \frac{μ_{o} i_{1}}{2 (2 π \times 10^{- 2})} = \frac{μ_{o} \times 3 \times 10^{2}}{4 π}$
$B_{2} = \frac{μ_{o} i_{2}}{2 (2 π \times 10^{- 2})} = \frac{μ_{o} \times 4 \times 10^{2}}{4 π}$
$B = \sqrt{B_{1}^{2} + B_{2}^{2}} = \frac{μ_{0}}{4 π} \times 5 \times 10^{2} = 10^{- 7} \times 5 \times 10^{2}$
$B = 5 \times 10^{- 5}$

AIIMS-2008

Moving Charges & Magnetism

153370 When a positively charged particle enters a uniform magnetic field with uniform velocity, its trajectory can be
1.a straight line 2. a circle 3 . a helix

1 (1) only

2 (1) or (2)

3 (1) or (3)

4 Any one of (1), (2) and (3)

Explanation:

D As we know that, magnetic force on a moving charge in a magnetic field is,
$F = qvB$
- When the particle moves along the direction of magnetic field then,
$F = 0$
Thus, the particle moves in a straight line along the incident direction.
- If the particle moves perpendicular to the direction of magnetic field,
Thus, the trajectory of the particle is circular.
- If the direction of velocity has both parallel and perpendicular component to the direction of magnetic field, the parallel component tends to move along the direction of magnetic field. While the perpendicular component moves the particle in a circle, thus the trajectory of the particle is helix.

EAMCET-2006

Moving Charges & Magnetism

153378 A square coil of side a carries a current $I$ . The magnetic field at the centre of the coil is

1

\frac{μ_{0} I}{a π}

2

\frac{\sqrt{2} μ_{0} I}{a π}

3

\frac{μ_{0} I}{\sqrt{2} a π}

4

\frac{2 \sqrt{2} μ_{0} I}{a π}

Explanation:

D $B_{total} = 4 B_{side}$
The magnetic field at the centre of the coil due to each side is in the same direction and are equal to each other,
$B_{total} = 4 \times \frac{μ_{o} I}{4 π (\frac{a}{2})} [\sin \frac{π}{4} + \sin \frac{π}{4}]$
$B_{total} = \frac{2 \sqrt{2} μ_{o} I}{a π}$

AIIMS-2012

Moving Charges & Magnetism

153379 Circular loop of a wire and a long straight wire carry currents $I_{c}$ and $I_{e}$ , respectively as shown in figure. Assuming that these are placed in the same plane, the magnetic fields will be zero at the centre of the loop when the separation $H$ is:

1

\frac{I_{e} R}{I_{c} π}

2

\frac{I_{c} R}{I_{e} π}

3

\frac{π I_{c}}{I_{e} R}

4

\frac{I_{e} π}{I_{c} R}

Explanation:

A Magnetic field due to straight wire $= \frac{μ_{o} I_{e}}{2 π H}$ Magnetic field due to circular wire $= \frac{μ_{o} I_{c}}{2 R}$
Now, $\frac{μ_{o} I_{e}}{2 π H} = \frac{μ_{o} I_{c}}{2 R}$
$H = \frac{{RI}_{e}}{π I_{c}}$

AIIMS-2006

Moving Charges & Magnetism

153381 Two concentric coils each of radius equal to $2 π$ cm are placed at right angles to each other. 3 ampere and 4 ampere are the currents flowing in each coil respectively. The magnetic induction in Weber $/ m^{2}$ at the centre of the coils will be.
$(μ_{0} = 4 π \times 10^{- 7} Wb / A . m)$

1

10^{- 5}

2

12 \times 10^{- 5}

3

7 \times 10^{- 5}

4

5 \times 10^{- 5}

Explanation:

D Magnetic field due to circular coil,
$I_{1} = 3 A, I_{2} = 4 A$
Where $r = 2 π cm = 2 π \times 10^{- 2} m$
$B_{1} = \frac{μ_{o} i_{1}}{2 r} = \frac{μ_{o} i_{1}}{2 (2 π \times 10^{- 2})} = \frac{μ_{o} \times 3 \times 10^{2}}{4 π}$
$B_{2} = \frac{μ_{o} i_{2}}{2 (2 π \times 10^{- 2})} = \frac{μ_{o} \times 4 \times 10^{2}}{4 π}$
$B = \sqrt{B_{1}^{2} + B_{2}^{2}} = \frac{μ_{0}}{4 π} \times 5 \times 10^{2} = 10^{- 7} \times 5 \times 10^{2}$
$B = 5 \times 10^{- 5}$

AIIMS-2008

Moving Charges & Magnetism

153370 When a positively charged particle enters a uniform magnetic field with uniform velocity, its trajectory can be
1.a straight line 2. a circle 3 . a helix

1 (1) only

2 (1) or (2)

3 (1) or (3)

4 Any one of (1), (2) and (3)

Explanation:

D As we know that, magnetic force on a moving charge in a magnetic field is,
$F = qvB$
- When the particle moves along the direction of magnetic field then,
$F = 0$
Thus, the particle moves in a straight line along the incident direction.
- If the particle moves perpendicular to the direction of magnetic field,
Thus, the trajectory of the particle is circular.
- If the direction of velocity has both parallel and perpendicular component to the direction of magnetic field, the parallel component tends to move along the direction of magnetic field. While the perpendicular component moves the particle in a circle, thus the trajectory of the particle is helix.

EAMCET-2006

Moving Charges & Magnetism

153378 A square coil of side a carries a current $I$ . The magnetic field at the centre of the coil is

1

\frac{μ_{0} I}{a π}

2

\frac{\sqrt{2} μ_{0} I}{a π}

3

\frac{μ_{0} I}{\sqrt{2} a π}

4

\frac{2 \sqrt{2} μ_{0} I}{a π}

Explanation:

D $B_{total} = 4 B_{side}$
The magnetic field at the centre of the coil due to each side is in the same direction and are equal to each other,
$B_{total} = 4 \times \frac{μ_{o} I}{4 π (\frac{a}{2})} [\sin \frac{π}{4} + \sin \frac{π}{4}]$
$B_{total} = \frac{2 \sqrt{2} μ_{o} I}{a π}$

AIIMS-2012

Moving Charges & Magnetism

153379 Circular loop of a wire and a long straight wire carry currents $I_{c}$ and $I_{e}$ , respectively as shown in figure. Assuming that these are placed in the same plane, the magnetic fields will be zero at the centre of the loop when the separation $H$ is:

1

\frac{I_{e} R}{I_{c} π}

2

\frac{I_{c} R}{I_{e} π}

3

\frac{π I_{c}}{I_{e} R}

4

\frac{I_{e} π}{I_{c} R}

Explanation:

A Magnetic field due to straight wire $= \frac{μ_{o} I_{e}}{2 π H}$ Magnetic field due to circular wire $= \frac{μ_{o} I_{c}}{2 R}$
Now, $\frac{μ_{o} I_{e}}{2 π H} = \frac{μ_{o} I_{c}}{2 R}$
$H = \frac{{RI}_{e}}{π I_{c}}$

AIIMS-2006

Moving Charges & Magnetism

153381 Two concentric coils each of radius equal to $2 π$ cm are placed at right angles to each other. 3 ampere and 4 ampere are the currents flowing in each coil respectively. The magnetic induction in Weber $/ m^{2}$ at the centre of the coils will be.
$(μ_{0} = 4 π \times 10^{- 7} Wb / A . m)$

1

10^{- 5}

2

12 \times 10^{- 5}

3

7 \times 10^{- 5}

4

5 \times 10^{- 5}

Explanation:

D Magnetic field due to circular coil,
$I_{1} = 3 A, I_{2} = 4 A$
Where $r = 2 π cm = 2 π \times 10^{- 2} m$
$B_{1} = \frac{μ_{o} i_{1}}{2 r} = \frac{μ_{o} i_{1}}{2 (2 π \times 10^{- 2})} = \frac{μ_{o} \times 3 \times 10^{2}}{4 π}$
$B_{2} = \frac{μ_{o} i_{2}}{2 (2 π \times 10^{- 2})} = \frac{μ_{o} \times 4 \times 10^{2}}{4 π}$
$B = \sqrt{B_{1}^{2} + B_{2}^{2}} = \frac{μ_{0}}{4 π} \times 5 \times 10^{2} = 10^{- 7} \times 5 \times 10^{2}$
$B = 5 \times 10^{- 5}$

AIIMS-2008

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