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Moving Charges & Magnetism

153155 If $B_{1}$ is the magnetic field induction at a point on the axis of a circular coil of radius $R$ situated at a distance $R \sqrt{3}$ and $B_{2}$ is the magnetic field at the centre of the coil, then the ratio of $\frac{B_{1}}{B_{2}}$ is equal to

1

\frac{1}{3}

2

\frac{1}{8}

3

\frac{1}{4}

4

\frac{1}{2}

Explanation:

B Magnetic field at centre of will
$B_{2} = \frac{μ_{0} I}{2 R}$
At, $r = R \sqrt{3}$
$B_{1} = \frac{μ_{o} {IR}^{2}}{2 {(R^{2} + r^{2})}^{3 / 2}}$
$B_{1} = \frac{μ_{0} I R^{2}}{2 {(R^{2} + 3 R^{2})}^{3 / 2}}$
$B_{1} = \frac{μ_{o} {IR}^{2}}{16 R^{3}}$
$\frac{B_{1}}{B_{2}} = \frac{1}{8}$

AP EAMCET (18.09.2020) Shift-II

Moving Charges & Magnetism

153158 A straight wire carrying current $I$ is turned into a circular loop. If the magnitude of magnetic moment associated with it in MKS unit is $M$ , the length of wire will be

1

4 π IM

2

\sqrt{\frac{4 π M}{I}}

3

\sqrt{\frac{4 π I}{M}}

4

\frac{M π}{4 I}

Explanation:

B Magnetic moment of straight current carrying wire-
$M = I A$
$M = I (π r^{2})$
Where, $r =$ radius of loop
When wire of length $L$ is turned into loop of radius $r$ then
$L = 2 π r$
$r = \frac{L}{2 π}$
Substituting the value of $r$ in equation (1)
$M = I π {(\frac{L}{2 π})}^{2}$
$L^{2} = \frac{4 π M}{I}$
$L = \sqrt{\frac{4 π M}{I}}$

AP EAMCET (18.09.2020) Shift-II

Moving Charges & Magnetism

153160 A horizontal overhead power line carries a current of $90 A$ in east to west direction. What is the magnitude and direction of the magnetic field due to the current, $1.5 m$ below the line?

1

1.2 \times 10^{- 5} T

towards south

2

1.2 \times 10^{- 5} T

towards north

3

1.2 \times 10^{- 5} T

towards east

4

1.2 \times 10^{- 5} T

towards west

Explanation:

A Magnetic field due to infinite wire,
$B = \frac{μ_{o} i}{2 π r}$
$i = 90 A, r = 1.5 m$
$B = \frac{4 π \times 10^{- 7} \times 90}{2 π \times 1.5}$
$B = 1.2 \times 10^{- 5} tesla$
Direction of magnetic field is finding out by right hand rule or palm rule which is towards south.

COMEDK 2020

Moving Charges & Magnetism

153162 Consider a long straight conducting wire. The magnetic field is determined as $B$ at a distance of $5 cm$ from the wire. The field at $40 cm$ from the wire would be:

1 4B

2

\frac{B}{2}

3

\frac{B}{8}

4

B

Explanation:

C In a long straight wire, Given,
$B_{1} = B$
$r_{1} = 5 cm$
$r_{2} = 40 cm$
$B = \frac{μ_{0} I}{2 π r}$
$B \propto \frac{1}{r}$
$\frac{B_{2}}{B_{1}} = \frac{r_{1}}{r_{2}} = \frac{5}{40}$
According to the question,
$∴ \frac{B_{2}}{B} = \frac{1}{8}$
$B_{2} = \frac{B}{8}$

TS EAMCET 29.09.2020

Moving Charges & Magnetism

153155 If $B_{1}$ is the magnetic field induction at a point on the axis of a circular coil of radius $R$ situated at a distance $R \sqrt{3}$ and $B_{2}$ is the magnetic field at the centre of the coil, then the ratio of $\frac{B_{1}}{B_{2}}$ is equal to

1

\frac{1}{3}

2

\frac{1}{8}

3

\frac{1}{4}

4

\frac{1}{2}

Explanation:

B Magnetic field at centre of will
$B_{2} = \frac{μ_{0} I}{2 R}$
At, $r = R \sqrt{3}$
$B_{1} = \frac{μ_{o} {IR}^{2}}{2 {(R^{2} + r^{2})}^{3 / 2}}$
$B_{1} = \frac{μ_{0} I R^{2}}{2 {(R^{2} + 3 R^{2})}^{3 / 2}}$
$B_{1} = \frac{μ_{o} {IR}^{2}}{16 R^{3}}$
$\frac{B_{1}}{B_{2}} = \frac{1}{8}$

AP EAMCET (18.09.2020) Shift-II

Moving Charges & Magnetism

153158 A straight wire carrying current $I$ is turned into a circular loop. If the magnitude of magnetic moment associated with it in MKS unit is $M$ , the length of wire will be

1

4 π IM

2

\sqrt{\frac{4 π M}{I}}

3

\sqrt{\frac{4 π I}{M}}

4

\frac{M π}{4 I}

Explanation:

B Magnetic moment of straight current carrying wire-
$M = I A$
$M = I (π r^{2})$
Where, $r =$ radius of loop
When wire of length $L$ is turned into loop of radius $r$ then
$L = 2 π r$
$r = \frac{L}{2 π}$
Substituting the value of $r$ in equation (1)
$M = I π {(\frac{L}{2 π})}^{2}$
$L^{2} = \frac{4 π M}{I}$
$L = \sqrt{\frac{4 π M}{I}}$

AP EAMCET (18.09.2020) Shift-II

Moving Charges & Magnetism

153160 A horizontal overhead power line carries a current of $90 A$ in east to west direction. What is the magnitude and direction of the magnetic field due to the current, $1.5 m$ below the line?

1

1.2 \times 10^{- 5} T

towards south

2

1.2 \times 10^{- 5} T

towards north

3

1.2 \times 10^{- 5} T

towards east

4

1.2 \times 10^{- 5} T

towards west

Explanation:

A Magnetic field due to infinite wire,
$B = \frac{μ_{o} i}{2 π r}$
$i = 90 A, r = 1.5 m$
$B = \frac{4 π \times 10^{- 7} \times 90}{2 π \times 1.5}$
$B = 1.2 \times 10^{- 5} tesla$
Direction of magnetic field is finding out by right hand rule or palm rule which is towards south.

COMEDK 2020

Moving Charges & Magnetism

153162 Consider a long straight conducting wire. The magnetic field is determined as $B$ at a distance of $5 cm$ from the wire. The field at $40 cm$ from the wire would be:

1 4B

2

\frac{B}{2}

3

\frac{B}{8}

4

B

Explanation:

C In a long straight wire, Given,
$B_{1} = B$
$r_{1} = 5 cm$
$r_{2} = 40 cm$
$B = \frac{μ_{0} I}{2 π r}$
$B \propto \frac{1}{r}$
$\frac{B_{2}}{B_{1}} = \frac{r_{1}}{r_{2}} = \frac{5}{40}$
According to the question,
$∴ \frac{B_{2}}{B} = \frac{1}{8}$
$B_{2} = \frac{B}{8}$

TS EAMCET 29.09.2020

Moving Charges & Magnetism

153155 If $B_{1}$ is the magnetic field induction at a point on the axis of a circular coil of radius $R$ situated at a distance $R \sqrt{3}$ and $B_{2}$ is the magnetic field at the centre of the coil, then the ratio of $\frac{B_{1}}{B_{2}}$ is equal to

1

\frac{1}{3}

2

\frac{1}{8}

3

\frac{1}{4}

4

\frac{1}{2}

Explanation:

B Magnetic field at centre of will
$B_{2} = \frac{μ_{0} I}{2 R}$
At, $r = R \sqrt{3}$
$B_{1} = \frac{μ_{o} {IR}^{2}}{2 {(R^{2} + r^{2})}^{3 / 2}}$
$B_{1} = \frac{μ_{0} I R^{2}}{2 {(R^{2} + 3 R^{2})}^{3 / 2}}$
$B_{1} = \frac{μ_{o} {IR}^{2}}{16 R^{3}}$
$\frac{B_{1}}{B_{2}} = \frac{1}{8}$

AP EAMCET (18.09.2020) Shift-II

Moving Charges & Magnetism

153158 A straight wire carrying current $I$ is turned into a circular loop. If the magnitude of magnetic moment associated with it in MKS unit is $M$ , the length of wire will be

1

4 π IM

2

\sqrt{\frac{4 π M}{I}}

3

\sqrt{\frac{4 π I}{M}}

4

\frac{M π}{4 I}

Explanation:

B Magnetic moment of straight current carrying wire-
$M = I A$
$M = I (π r^{2})$
Where, $r =$ radius of loop
When wire of length $L$ is turned into loop of radius $r$ then
$L = 2 π r$
$r = \frac{L}{2 π}$
Substituting the value of $r$ in equation (1)
$M = I π {(\frac{L}{2 π})}^{2}$
$L^{2} = \frac{4 π M}{I}$
$L = \sqrt{\frac{4 π M}{I}}$

AP EAMCET (18.09.2020) Shift-II

Moving Charges & Magnetism

153160 A horizontal overhead power line carries a current of $90 A$ in east to west direction. What is the magnitude and direction of the magnetic field due to the current, $1.5 m$ below the line?

1

1.2 \times 10^{- 5} T

towards south

2

1.2 \times 10^{- 5} T

towards north

3

1.2 \times 10^{- 5} T

towards east

4

1.2 \times 10^{- 5} T

towards west

Explanation:

A Magnetic field due to infinite wire,
$B = \frac{μ_{o} i}{2 π r}$
$i = 90 A, r = 1.5 m$
$B = \frac{4 π \times 10^{- 7} \times 90}{2 π \times 1.5}$
$B = 1.2 \times 10^{- 5} tesla$
Direction of magnetic field is finding out by right hand rule or palm rule which is towards south.

COMEDK 2020

Moving Charges & Magnetism

153162 Consider a long straight conducting wire. The magnetic field is determined as $B$ at a distance of $5 cm$ from the wire. The field at $40 cm$ from the wire would be:

1 4B

2

\frac{B}{2}

3

\frac{B}{8}

4

B

Explanation:

C In a long straight wire, Given,
$B_{1} = B$
$r_{1} = 5 cm$
$r_{2} = 40 cm$
$B = \frac{μ_{0} I}{2 π r}$
$B \propto \frac{1}{r}$
$\frac{B_{2}}{B_{1}} = \frac{r_{1}}{r_{2}} = \frac{5}{40}$
According to the question,
$∴ \frac{B_{2}}{B} = \frac{1}{8}$
$B_{2} = \frac{B}{8}$

TS EAMCET 29.09.2020

Moving Charges & Magnetism

153155 If $B_{1}$ is the magnetic field induction at a point on the axis of a circular coil of radius $R$ situated at a distance $R \sqrt{3}$ and $B_{2}$ is the magnetic field at the centre of the coil, then the ratio of $\frac{B_{1}}{B_{2}}$ is equal to

1

\frac{1}{3}

2

\frac{1}{8}

3

\frac{1}{4}

4

\frac{1}{2}

Explanation:

B Magnetic field at centre of will
$B_{2} = \frac{μ_{0} I}{2 R}$
At, $r = R \sqrt{3}$
$B_{1} = \frac{μ_{o} {IR}^{2}}{2 {(R^{2} + r^{2})}^{3 / 2}}$
$B_{1} = \frac{μ_{0} I R^{2}}{2 {(R^{2} + 3 R^{2})}^{3 / 2}}$
$B_{1} = \frac{μ_{o} {IR}^{2}}{16 R^{3}}$
$\frac{B_{1}}{B_{2}} = \frac{1}{8}$

AP EAMCET (18.09.2020) Shift-II

Moving Charges & Magnetism

153158 A straight wire carrying current $I$ is turned into a circular loop. If the magnitude of magnetic moment associated with it in MKS unit is $M$ , the length of wire will be

1

4 π IM

2

\sqrt{\frac{4 π M}{I}}

3

\sqrt{\frac{4 π I}{M}}

4

\frac{M π}{4 I}

Explanation:

B Magnetic moment of straight current carrying wire-
$M = I A$
$M = I (π r^{2})$
Where, $r =$ radius of loop
When wire of length $L$ is turned into loop of radius $r$ then
$L = 2 π r$
$r = \frac{L}{2 π}$
Substituting the value of $r$ in equation (1)
$M = I π {(\frac{L}{2 π})}^{2}$
$L^{2} = \frac{4 π M}{I}$
$L = \sqrt{\frac{4 π M}{I}}$

AP EAMCET (18.09.2020) Shift-II

Moving Charges & Magnetism

153160 A horizontal overhead power line carries a current of $90 A$ in east to west direction. What is the magnitude and direction of the magnetic field due to the current, $1.5 m$ below the line?

1

1.2 \times 10^{- 5} T

towards south

2

1.2 \times 10^{- 5} T

towards north

3

1.2 \times 10^{- 5} T

towards east

4

1.2 \times 10^{- 5} T

towards west

Explanation:

A Magnetic field due to infinite wire,
$B = \frac{μ_{o} i}{2 π r}$
$i = 90 A, r = 1.5 m$
$B = \frac{4 π \times 10^{- 7} \times 90}{2 π \times 1.5}$
$B = 1.2 \times 10^{- 5} tesla$
Direction of magnetic field is finding out by right hand rule or palm rule which is towards south.

COMEDK 2020

Moving Charges & Magnetism

153162 Consider a long straight conducting wire. The magnetic field is determined as $B$ at a distance of $5 cm$ from the wire. The field at $40 cm$ from the wire would be:

1 4B

2

\frac{B}{2}

3

\frac{B}{8}

4

B

Explanation:

C In a long straight wire, Given,
$B_{1} = B$
$r_{1} = 5 cm$
$r_{2} = 40 cm$
$B = \frac{μ_{0} I}{2 π r}$
$B \propto \frac{1}{r}$
$\frac{B_{2}}{B_{1}} = \frac{r_{1}}{r_{2}} = \frac{5}{40}$
According to the question,
$∴ \frac{B_{2}}{B} = \frac{1}{8}$
$B_{2} = \frac{B}{8}$

TS EAMCET 29.09.2020

Question Bank Series

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