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Moving Charges & Magnetism

153373 The lines of force of a uniform magnetic field:

1 must be convergent.

2 must be divergent.

3 must be parallel to each other.

4 intersect.

NDA (I) 2011

Moving Charges & Magnetism

153374 A positively charged particle projected towards west is deflected towards north by a magnetic field. The direction of the magnetic field is

1 towards South

2 towards east

3 in downward direction

4 in upward direction

NDA (I) 2013

Moving Charges & Magnetism

153377 If we double the radius of a coil keeping the current through it unchanged, then the magnetic field at any point at a large distance from the centre becomes approximately

1 double

2 three times

3 four times

4 one-fourth

Explanation:

C Magnetic field at an axial point $d >> R$ is
$B = \frac{μ_{0} I R^{2}}{2 {(d^{2} + R^{2})}^{3 / 2}}$
$B = (\frac{μ_{0} I}{2 d^{3}}) R^{2}$
Where $I$ and $d$ is constant, we get $B \propto R^{2}$
Thus, when radius is doubled, the magnetic field becomes four times hence option (c) is correct.

AIIMS-2014

Moving Charges & Magnetism

153380 What is the magnetic field at a distance $R$ from a coil of radius $r$ carrying current $I$ ?

1

\frac{μ_{0} {IR}^{2}}{2 {(R^{2} + r^{2})}^{\frac{3}{2}}}

2

\frac{μ_{0} {Ir}^{2}}{2 {(R^{2} + r^{2})}^{\frac{3}{2}}}

3

\frac{μ_{0} I}{2 r}

4

\frac{μ_{0} I}{2 R}

Explanation:

B
This can be resolved into two component, one along the axis OP and other PS, which is perpendicular to OP. So, the total magnetic field at a point which is at a distance $x$ away from the axis of a circular coil of radius $r$ is given by.
$B_{x} = \frac{μ_{0} {Ir}^{2}}{2 {(R^{2} + r^{2})}^{3 / 2}}$

AIIMS-2007

Moving Charges & Magnetism

153373 The lines of force of a uniform magnetic field:

1 must be convergent.

2 must be divergent.

3 must be parallel to each other.

4 intersect.

NDA (I) 2011

Moving Charges & Magnetism

153374 A positively charged particle projected towards west is deflected towards north by a magnetic field. The direction of the magnetic field is

1 towards South

2 towards east

3 in downward direction

4 in upward direction

NDA (I) 2013

Moving Charges & Magnetism

153377 If we double the radius of a coil keeping the current through it unchanged, then the magnetic field at any point at a large distance from the centre becomes approximately

1 double

2 three times

3 four times

4 one-fourth

Explanation:

C Magnetic field at an axial point $d >> R$ is
$B = \frac{μ_{0} I R^{2}}{2 {(d^{2} + R^{2})}^{3 / 2}}$
$B = (\frac{μ_{0} I}{2 d^{3}}) R^{2}$
Where $I$ and $d$ is constant, we get $B \propto R^{2}$
Thus, when radius is doubled, the magnetic field becomes four times hence option (c) is correct.

AIIMS-2014

Moving Charges & Magnetism

153380 What is the magnetic field at a distance $R$ from a coil of radius $r$ carrying current $I$ ?

1

\frac{μ_{0} {IR}^{2}}{2 {(R^{2} + r^{2})}^{\frac{3}{2}}}

2

\frac{μ_{0} {Ir}^{2}}{2 {(R^{2} + r^{2})}^{\frac{3}{2}}}

3

\frac{μ_{0} I}{2 r}

4

\frac{μ_{0} I}{2 R}

Explanation:

B
This can be resolved into two component, one along the axis OP and other PS, which is perpendicular to OP. So, the total magnetic field at a point which is at a distance $x$ away from the axis of a circular coil of radius $r$ is given by.
$B_{x} = \frac{μ_{0} {Ir}^{2}}{2 {(R^{2} + r^{2})}^{3 / 2}}$

AIIMS-2007

Moving Charges & Magnetism

153373 The lines of force of a uniform magnetic field:

1 must be convergent.

2 must be divergent.

3 must be parallel to each other.

4 intersect.

NDA (I) 2011

Moving Charges & Magnetism

153374 A positively charged particle projected towards west is deflected towards north by a magnetic field. The direction of the magnetic field is

1 towards South

2 towards east

3 in downward direction

4 in upward direction

NDA (I) 2013

Moving Charges & Magnetism

153377 If we double the radius of a coil keeping the current through it unchanged, then the magnetic field at any point at a large distance from the centre becomes approximately

1 double

2 three times

3 four times

4 one-fourth

Explanation:

C Magnetic field at an axial point $d >> R$ is
$B = \frac{μ_{0} I R^{2}}{2 {(d^{2} + R^{2})}^{3 / 2}}$
$B = (\frac{μ_{0} I}{2 d^{3}}) R^{2}$
Where $I$ and $d$ is constant, we get $B \propto R^{2}$
Thus, when radius is doubled, the magnetic field becomes four times hence option (c) is correct.

AIIMS-2014

Moving Charges & Magnetism

153380 What is the magnetic field at a distance $R$ from a coil of radius $r$ carrying current $I$ ?

1

\frac{μ_{0} {IR}^{2}}{2 {(R^{2} + r^{2})}^{\frac{3}{2}}}

2

\frac{μ_{0} {Ir}^{2}}{2 {(R^{2} + r^{2})}^{\frac{3}{2}}}

3

\frac{μ_{0} I}{2 r}

4

\frac{μ_{0} I}{2 R}

Explanation:

B
This can be resolved into two component, one along the axis OP and other PS, which is perpendicular to OP. So, the total magnetic field at a point which is at a distance $x$ away from the axis of a circular coil of radius $r$ is given by.
$B_{x} = \frac{μ_{0} {Ir}^{2}}{2 {(R^{2} + r^{2})}^{3 / 2}}$

AIIMS-2007

Moving Charges & Magnetism

153373 The lines of force of a uniform magnetic field:

1 must be convergent.

2 must be divergent.

3 must be parallel to each other.

4 intersect.

NDA (I) 2011

Moving Charges & Magnetism

153374 A positively charged particle projected towards west is deflected towards north by a magnetic field. The direction of the magnetic field is

1 towards South

2 towards east

3 in downward direction

4 in upward direction

NDA (I) 2013

Moving Charges & Magnetism

153377 If we double the radius of a coil keeping the current through it unchanged, then the magnetic field at any point at a large distance from the centre becomes approximately

1 double

2 three times

3 four times

4 one-fourth

Explanation:

C Magnetic field at an axial point $d >> R$ is
$B = \frac{μ_{0} I R^{2}}{2 {(d^{2} + R^{2})}^{3 / 2}}$
$B = (\frac{μ_{0} I}{2 d^{3}}) R^{2}$
Where $I$ and $d$ is constant, we get $B \propto R^{2}$
Thus, when radius is doubled, the magnetic field becomes four times hence option (c) is correct.

AIIMS-2014

Moving Charges & Magnetism

153380 What is the magnetic field at a distance $R$ from a coil of radius $r$ carrying current $I$ ?

1

\frac{μ_{0} {IR}^{2}}{2 {(R^{2} + r^{2})}^{\frac{3}{2}}}

2

\frac{μ_{0} {Ir}^{2}}{2 {(R^{2} + r^{2})}^{\frac{3}{2}}}

3

\frac{μ_{0} I}{2 r}

4

\frac{μ_{0} I}{2 R}

Explanation:

B
This can be resolved into two component, one along the axis OP and other PS, which is perpendicular to OP. So, the total magnetic field at a point which is at a distance $x$ away from the axis of a circular coil of radius $r$ is given by.
$B_{x} = \frac{μ_{0} {Ir}^{2}}{2 {(R^{2} + r^{2})}^{3 / 2}}$

AIIMS-2007

Question Bank Series

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Question Bank Series

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