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Moving Charges & Magnetism

153910 Two wires of same length and material are used to form a square loop and a circular loop respectively. If same current is passed through both loops then the ratio of magnetic moment of square loop to that of circular loop is

1

π

2

\frac{π}{4}

3

2 π

4

\frac{π}{2}

Explanation:

B Let the length of the wire L
Then, side o square, $L = 4 a \Rightarrow a = L / 4$
Then, area $(A_{1}) = (L / 4)^{2}$
Radius of circular loop,
$2 π r = L$
$r = \frac{L}{2 π}$
Then, area $(A_{2}) = 2 π {(\frac{L}{2 π})}^{2}$
We know,
Magnetic moment $(M) = IA$
${(M_{1})}_{square} = {IA}_{1}$
$= I \times {(\frac{L}{4})}^{2}$
And ${(M_{2})}_{circular loop} = {IA}_{2}$
$= I \times 2 π {(\frac{L}{2 π})}^{2}$
From equation (i) and equation (ii), we get -
$\frac{{(M_{1})}_{square}}{{(M_{2})}_{circular loop}} = \frac{I \times {(\frac{L}{4})}^{2}}{I \times 2 π {(\frac{L}{2 π})}^{2}} = \frac{π}{4}$

MHT-CET 2020

Moving Charges & Magnetism

153912 What is the magnetic moment of a current carrying circular coil if the radius of the circular coil is ' $R$ ' and magnetic induction at the center is ' $B$ '?

1

\frac{2 π {BR}^{3}}{μ_{0}}

2

\frac{π {BR}^{3}}{μ_{0}}

3

\frac{π B^{2} R^{2}}{μ_{0}}

4

\frac{π B^{2} R^{3}}{μ_{0}}

Explanation:

A Magnetic moment $(M) = IA = I (π R^{2})$
Magnetic field of a circular coil,
$B = \frac{μ_{0} I}{2 R}$
$I = \frac{2 B R}{μ_{0}}$
From equation (i) and (ii)
$M = \frac{2 B R}{μ_{0}} \times π R^{2}$
$M = \frac{2 π B R^{3}}{μ_{0}}$

MHT-CET 2020

Moving Charges & Magnetism

153913 A straight wire carrying a current ' $I$ ' is turned into a circular loop. If the magnetic moment associated with it is ' $M$ ', the length of the wire will be

1

\sqrt{\frac{π M}{I}}

2

2 \sqrt{\frac{π M}{I}}

3

\frac{2 π M}{I}

4

\frac{π M}{I}

Explanation:

B $M = IA = I (π r^{2})$
When wire of length $L$ is turned into loop radius-
or $r = \frac{L}{2 π}$
Putting the value of ' $r$ ' in equation (i), we get-
$M = I π {(\frac{L}{2 π})}^{2}$
$L^{2} = \frac{4 π M}{I}$
$L = 2 \sqrt{\frac{π M}{I}}$

MHT-CET 2020

Moving Charges & Magnetism

153914 The ratio of magnetic field at the centre of a current carrying circular coil to its magnetic moment is $x$ , if the current and the radius both are doubled. The new ratio will become :

1

2 x

2

4 x

3

\frac{x}{4}

4

\frac{x}{8}

Explanation:

D Magnetic field of the centre of a circular loop of radius $R$ carrying current $I$ .
$B = \frac{μ_{o} \cdot 2 π I}{4 π R} = \frac{μ_{o} I}{2 R}$
Magnetic moment $(M) = IA$
$M = I (π \cdot R^{2}) (∵ A = π R^{2})$
$\frac{B}{M} = \frac{μ_{o} \cdot I}{2 R} \times \frac{1}{π R^{2} I} = \frac{μ_{o}}{2 π R^{3}} = x$
When both the current and radius are doubled the ratio becomes.
$\frac{B^{'}}{M^{'}} = \frac{μ_{o}}{2 π (2 R)^{3}}$
$\frac{B^{'}}{M^{'}} = \frac{1}{8} (\frac{μ_{o}}{2 π R^{3}}) (∵ x = \frac{μ_{0}}{2 π R^{3}})$
$\frac{B^{'}}{M^{'}} = \frac{x}{8}$

Karnataka CET-2020

Moving Charges & Magnetism

153910 Two wires of same length and material are used to form a square loop and a circular loop respectively. If same current is passed through both loops then the ratio of magnetic moment of square loop to that of circular loop is

1

π

2

\frac{π}{4}

3

2 π

4

\frac{π}{2}

Explanation:

B Let the length of the wire L
Then, side o square, $L = 4 a \Rightarrow a = L / 4$
Then, area $(A_{1}) = (L / 4)^{2}$
Radius of circular loop,
$2 π r = L$
$r = \frac{L}{2 π}$
Then, area $(A_{2}) = 2 π {(\frac{L}{2 π})}^{2}$
We know,
Magnetic moment $(M) = IA$
${(M_{1})}_{square} = {IA}_{1}$
$= I \times {(\frac{L}{4})}^{2}$
And ${(M_{2})}_{circular loop} = {IA}_{2}$
$= I \times 2 π {(\frac{L}{2 π})}^{2}$
From equation (i) and equation (ii), we get -
$\frac{{(M_{1})}_{square}}{{(M_{2})}_{circular loop}} = \frac{I \times {(\frac{L}{4})}^{2}}{I \times 2 π {(\frac{L}{2 π})}^{2}} = \frac{π}{4}$

MHT-CET 2020

Moving Charges & Magnetism

153912 What is the magnetic moment of a current carrying circular coil if the radius of the circular coil is ' $R$ ' and magnetic induction at the center is ' $B$ '?

1

\frac{2 π {BR}^{3}}{μ_{0}}

2

\frac{π {BR}^{3}}{μ_{0}}

3

\frac{π B^{2} R^{2}}{μ_{0}}

4

\frac{π B^{2} R^{3}}{μ_{0}}

Explanation:

A Magnetic moment $(M) = IA = I (π R^{2})$
Magnetic field of a circular coil,
$B = \frac{μ_{0} I}{2 R}$
$I = \frac{2 B R}{μ_{0}}$
From equation (i) and (ii)
$M = \frac{2 B R}{μ_{0}} \times π R^{2}$
$M = \frac{2 π B R^{3}}{μ_{0}}$

MHT-CET 2020

Moving Charges & Magnetism

153913 A straight wire carrying a current ' $I$ ' is turned into a circular loop. If the magnetic moment associated with it is ' $M$ ', the length of the wire will be

1

\sqrt{\frac{π M}{I}}

2

2 \sqrt{\frac{π M}{I}}

3

\frac{2 π M}{I}

4

\frac{π M}{I}

Explanation:

B $M = IA = I (π r^{2})$
When wire of length $L$ is turned into loop radius-
or $r = \frac{L}{2 π}$
Putting the value of ' $r$ ' in equation (i), we get-
$M = I π {(\frac{L}{2 π})}^{2}$
$L^{2} = \frac{4 π M}{I}$
$L = 2 \sqrt{\frac{π M}{I}}$

MHT-CET 2020

Moving Charges & Magnetism

153914 The ratio of magnetic field at the centre of a current carrying circular coil to its magnetic moment is $x$ , if the current and the radius both are doubled. The new ratio will become :

1

2 x

2

4 x

3

\frac{x}{4}

4

\frac{x}{8}

Explanation:

D Magnetic field of the centre of a circular loop of radius $R$ carrying current $I$ .
$B = \frac{μ_{o} \cdot 2 π I}{4 π R} = \frac{μ_{o} I}{2 R}$
Magnetic moment $(M) = IA$
$M = I (π \cdot R^{2}) (∵ A = π R^{2})$
$\frac{B}{M} = \frac{μ_{o} \cdot I}{2 R} \times \frac{1}{π R^{2} I} = \frac{μ_{o}}{2 π R^{3}} = x$
When both the current and radius are doubled the ratio becomes.
$\frac{B^{'}}{M^{'}} = \frac{μ_{o}}{2 π (2 R)^{3}}$
$\frac{B^{'}}{M^{'}} = \frac{1}{8} (\frac{μ_{o}}{2 π R^{3}}) (∵ x = \frac{μ_{0}}{2 π R^{3}})$
$\frac{B^{'}}{M^{'}} = \frac{x}{8}$

Karnataka CET-2020

Moving Charges & Magnetism

153910 Two wires of same length and material are used to form a square loop and a circular loop respectively. If same current is passed through both loops then the ratio of magnetic moment of square loop to that of circular loop is

1

π

2

\frac{π}{4}

3

2 π

4

\frac{π}{2}

Explanation:

B Let the length of the wire L
Then, side o square, $L = 4 a \Rightarrow a = L / 4$
Then, area $(A_{1}) = (L / 4)^{2}$
Radius of circular loop,
$2 π r = L$
$r = \frac{L}{2 π}$
Then, area $(A_{2}) = 2 π {(\frac{L}{2 π})}^{2}$
We know,
Magnetic moment $(M) = IA$
${(M_{1})}_{square} = {IA}_{1}$
$= I \times {(\frac{L}{4})}^{2}$
And ${(M_{2})}_{circular loop} = {IA}_{2}$
$= I \times 2 π {(\frac{L}{2 π})}^{2}$
From equation (i) and equation (ii), we get -
$\frac{{(M_{1})}_{square}}{{(M_{2})}_{circular loop}} = \frac{I \times {(\frac{L}{4})}^{2}}{I \times 2 π {(\frac{L}{2 π})}^{2}} = \frac{π}{4}$

MHT-CET 2020

Moving Charges & Magnetism

153912 What is the magnetic moment of a current carrying circular coil if the radius of the circular coil is ' $R$ ' and magnetic induction at the center is ' $B$ '?

1

\frac{2 π {BR}^{3}}{μ_{0}}

2

\frac{π {BR}^{3}}{μ_{0}}

3

\frac{π B^{2} R^{2}}{μ_{0}}

4

\frac{π B^{2} R^{3}}{μ_{0}}

Explanation:

A Magnetic moment $(M) = IA = I (π R^{2})$
Magnetic field of a circular coil,
$B = \frac{μ_{0} I}{2 R}$
$I = \frac{2 B R}{μ_{0}}$
From equation (i) and (ii)
$M = \frac{2 B R}{μ_{0}} \times π R^{2}$
$M = \frac{2 π B R^{3}}{μ_{0}}$

MHT-CET 2020

Moving Charges & Magnetism

153913 A straight wire carrying a current ' $I$ ' is turned into a circular loop. If the magnetic moment associated with it is ' $M$ ', the length of the wire will be

1

\sqrt{\frac{π M}{I}}

2

2 \sqrt{\frac{π M}{I}}

3

\frac{2 π M}{I}

4

\frac{π M}{I}

Explanation:

B $M = IA = I (π r^{2})$
When wire of length $L$ is turned into loop radius-
or $r = \frac{L}{2 π}$
Putting the value of ' $r$ ' in equation (i), we get-
$M = I π {(\frac{L}{2 π})}^{2}$
$L^{2} = \frac{4 π M}{I}$
$L = 2 \sqrt{\frac{π M}{I}}$

MHT-CET 2020

Moving Charges & Magnetism

153914 The ratio of magnetic field at the centre of a current carrying circular coil to its magnetic moment is $x$ , if the current and the radius both are doubled. The new ratio will become :

1

2 x

2

4 x

3

\frac{x}{4}

4

\frac{x}{8}

Explanation:

D Magnetic field of the centre of a circular loop of radius $R$ carrying current $I$ .
$B = \frac{μ_{o} \cdot 2 π I}{4 π R} = \frac{μ_{o} I}{2 R}$
Magnetic moment $(M) = IA$
$M = I (π \cdot R^{2}) (∵ A = π R^{2})$
$\frac{B}{M} = \frac{μ_{o} \cdot I}{2 R} \times \frac{1}{π R^{2} I} = \frac{μ_{o}}{2 π R^{3}} = x$
When both the current and radius are doubled the ratio becomes.
$\frac{B^{'}}{M^{'}} = \frac{μ_{o}}{2 π (2 R)^{3}}$
$\frac{B^{'}}{M^{'}} = \frac{1}{8} (\frac{μ_{o}}{2 π R^{3}}) (∵ x = \frac{μ_{0}}{2 π R^{3}})$
$\frac{B^{'}}{M^{'}} = \frac{x}{8}$

Karnataka CET-2020

Moving Charges & Magnetism

153910 Two wires of same length and material are used to form a square loop and a circular loop respectively. If same current is passed through both loops then the ratio of magnetic moment of square loop to that of circular loop is

1

π

2

\frac{π}{4}

3

2 π

4

\frac{π}{2}

Explanation:

B Let the length of the wire L
Then, side o square, $L = 4 a \Rightarrow a = L / 4$
Then, area $(A_{1}) = (L / 4)^{2}$
Radius of circular loop,
$2 π r = L$
$r = \frac{L}{2 π}$
Then, area $(A_{2}) = 2 π {(\frac{L}{2 π})}^{2}$
We know,
Magnetic moment $(M) = IA$
${(M_{1})}_{square} = {IA}_{1}$
$= I \times {(\frac{L}{4})}^{2}$
And ${(M_{2})}_{circular loop} = {IA}_{2}$
$= I \times 2 π {(\frac{L}{2 π})}^{2}$
From equation (i) and equation (ii), we get -
$\frac{{(M_{1})}_{square}}{{(M_{2})}_{circular loop}} = \frac{I \times {(\frac{L}{4})}^{2}}{I \times 2 π {(\frac{L}{2 π})}^{2}} = \frac{π}{4}$

MHT-CET 2020

Moving Charges & Magnetism

153912 What is the magnetic moment of a current carrying circular coil if the radius of the circular coil is ' $R$ ' and magnetic induction at the center is ' $B$ '?

1

\frac{2 π {BR}^{3}}{μ_{0}}

2

\frac{π {BR}^{3}}{μ_{0}}

3

\frac{π B^{2} R^{2}}{μ_{0}}

4

\frac{π B^{2} R^{3}}{μ_{0}}

Explanation:

A Magnetic moment $(M) = IA = I (π R^{2})$
Magnetic field of a circular coil,
$B = \frac{μ_{0} I}{2 R}$
$I = \frac{2 B R}{μ_{0}}$
From equation (i) and (ii)
$M = \frac{2 B R}{μ_{0}} \times π R^{2}$
$M = \frac{2 π B R^{3}}{μ_{0}}$

MHT-CET 2020

Moving Charges & Magnetism

153913 A straight wire carrying a current ' $I$ ' is turned into a circular loop. If the magnetic moment associated with it is ' $M$ ', the length of the wire will be

1

\sqrt{\frac{π M}{I}}

2

2 \sqrt{\frac{π M}{I}}

3

\frac{2 π M}{I}

4

\frac{π M}{I}

Explanation:

B $M = IA = I (π r^{2})$
When wire of length $L$ is turned into loop radius-
or $r = \frac{L}{2 π}$
Putting the value of ' $r$ ' in equation (i), we get-
$M = I π {(\frac{L}{2 π})}^{2}$
$L^{2} = \frac{4 π M}{I}$
$L = 2 \sqrt{\frac{π M}{I}}$

MHT-CET 2020

Moving Charges & Magnetism

153914 The ratio of magnetic field at the centre of a current carrying circular coil to its magnetic moment is $x$ , if the current and the radius both are doubled. The new ratio will become :

1

2 x

2

4 x

3

\frac{x}{4}

4

\frac{x}{8}

Explanation:

D Magnetic field of the centre of a circular loop of radius $R$ carrying current $I$ .
$B = \frac{μ_{o} \cdot 2 π I}{4 π R} = \frac{μ_{o} I}{2 R}$
Magnetic moment $(M) = IA$
$M = I (π \cdot R^{2}) (∵ A = π R^{2})$
$\frac{B}{M} = \frac{μ_{o} \cdot I}{2 R} \times \frac{1}{π R^{2} I} = \frac{μ_{o}}{2 π R^{3}} = x$
When both the current and radius are doubled the ratio becomes.
$\frac{B^{'}}{M^{'}} = \frac{μ_{o}}{2 π (2 R)^{3}}$
$\frac{B^{'}}{M^{'}} = \frac{1}{8} (\frac{μ_{o}}{2 π R^{3}}) (∵ x = \frac{μ_{0}}{2 π R^{3}})$
$\frac{B^{'}}{M^{'}} = \frac{x}{8}$

Karnataka CET-2020

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