QBManager

Moving Charges & Magnetism

153755 A wire carrying current ' $I$ ' along $x$ axis has length ' $ℓ$ ' and it is kept in a magnetic field $\vec{B} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) B \frac{W b}{m^{2}}$ . The magnitude of magnetic force acting on the wire is

1

\sqrt{11} I ℓ B

2

\sqrt{15} I ℓ B

3

\sqrt{13} I ℓ B

4

\sqrt{19} I ℓ B

Explanation:

C Given,
$\vec{B} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) B \frac{W b}{m^{2}}$
Current passing along $x$ -axis
$F = i l B$
$F = l (\hat{Ii}) B (\hat{i} + 2 \hat{j} - 3 \hat{k})$
$F = l IB (+ 2 \hat{j} - 3 \hat{k})$
$∣ F = l IB \sqrt{4 + 9}$
$| F | = \sqrt{13} I l B$

MHT-CET 2020

Moving Charges & Magnetism

153757 A wire of length ' $L$ ' carries current ' $I$ ' along $x$ axis, A magnetic field $B = B_{0} (\hat{i} - \hat{j} - \hat{k}) T$ acts on the wire. The magnitude of magnetic force acting on the wire is

1

\frac{{ILB}_{0}}{2}

2 I L B

_{0}

3

\sqrt{2} {ILB}_{0}

4

2 {ILB}_{0}

Explanation:

C Given, $B = B_{0} (\hat{i} - \hat{j} - \hat{k}) T$ and current $= I \hat{i}$
Force acting on the wire
$F = B I L$
$F = L (I \hat{i}) B_{0} (\hat{i} - \hat{j} - \hat{k})$
$F = B_{0} I L (+ \hat{j} - \hat{k})$
$| F | = B_{0} I L \sqrt{1 + 1} = \sqrt{2} B_{0} I L$
Hence, magnitude of magnetic force, $| F | = \sqrt{2} B_{0} IL$

MHT-CET 2020

Moving Charges & Magnetism

153758 A metal wire of length ' $L$ ' is bent to form a circular coil of number of turns ' $n$ '. The coil is placed in magnetic field ' $B$ ' and current is passed through the coil. The maximum torque acting on the coil is

1

\frac{B^{2} I L}{2 π n}

2

\frac{{BIL}^{2}}{2 π n}

3

\frac{{BIL}^{2}}{4 π n}

4

\frac{B^{2} I L}{4 π n}

Explanation:

C Given,
$L = 2 π rn$
$r = \frac{L}{2 π n}$
When current I passed through the loop the maximum torque acting on the loop is
$τ = IBAn$
$τ = IB π r^{2} n$
$τ = IB π \frac{L^{2}}{π^{2} n^{2}} \times n$
$τ = IB \frac{L^{2}}{4 π n}$

MHT-CET 2020

Moving Charges & Magnetism

153759 Given figure shows the north and south poles of a permanent magnet in which a coil of $n$ turns of cross-sectional area $A$ is resting, such that when a current $I$ is passed through the coil, the plane of the coil makes and angle $θ$ with respect to direction of magnetic field $B$ . If the plane of magnetic field and the coil are horizontal and vertical respectively, the torque on the coil will be

1

n I A B \cos θ

2

n I A B \sin θ

3 nIAB

4 None of the above, since the magnetic field is radial

Explanation:

A Plane of coil having angle $θ$ with the magnetic field.
$τ = MB \sin (90^{\circ} - θ)$
Magnetic moment $(M) = nIA$
From equation (i) and equation (ii), we get -
$τ = n I A B \cos θ$

AP EAMCET (17.09.2020) Shift-II

Moving Charges & Magnetism

153760 A coil of 50 turns carrying a current of $2 A$ in a magnetic field of $0.5 T$ . The torque acting on the coil is

1

0.4 Nm

clockwise

2

0.2 Nm

anticlockwise

3

0.4 Nm

anticlockwise

4

0.2 Nm

clockwise

5

0.8 Nm

anticlockwise

Explanation:

A Given, Number of turns on the coil, $N = 50$
Current carried by the coil, $I = 2 A$
Magnetic field, $B = 0.5 T$

According to the figure,
The sides $AB$ and $DC$ are along the field lines hence, the force on each side is zero.
The torque on each vertical wire is given as
$τ = B I N A \sin θ$
$τ = 0.5 \times 2 \times 50 \times 0.10 \times 0.08 \sin 90 (∵ \sin θ = 1)$
$τ = 50 \times 0.008$
$τ = 0.4 Nm (clockwise) .$

Kerala CEE 2020

Moving Charges & Magnetism

153755 A wire carrying current ' $I$ ' along $x$ axis has length ' $ℓ$ ' and it is kept in a magnetic field $\vec{B} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) B \frac{W b}{m^{2}}$ . The magnitude of magnetic force acting on the wire is

1

\sqrt{11} I ℓ B

2

\sqrt{15} I ℓ B

3

\sqrt{13} I ℓ B

4

\sqrt{19} I ℓ B

Explanation:

C Given,
$\vec{B} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) B \frac{W b}{m^{2}}$
Current passing along $x$ -axis
$F = i l B$
$F = l (\hat{Ii}) B (\hat{i} + 2 \hat{j} - 3 \hat{k})$
$F = l IB (+ 2 \hat{j} - 3 \hat{k})$
$∣ F = l IB \sqrt{4 + 9}$
$| F | = \sqrt{13} I l B$

MHT-CET 2020

Moving Charges & Magnetism

153757 A wire of length ' $L$ ' carries current ' $I$ ' along $x$ axis, A magnetic field $B = B_{0} (\hat{i} - \hat{j} - \hat{k}) T$ acts on the wire. The magnitude of magnetic force acting on the wire is

1

\frac{{ILB}_{0}}{2}

2 I L B

_{0}

3

\sqrt{2} {ILB}_{0}

4

2 {ILB}_{0}

Explanation:

C Given, $B = B_{0} (\hat{i} - \hat{j} - \hat{k}) T$ and current $= I \hat{i}$
Force acting on the wire
$F = B I L$
$F = L (I \hat{i}) B_{0} (\hat{i} - \hat{j} - \hat{k})$
$F = B_{0} I L (+ \hat{j} - \hat{k})$
$| F | = B_{0} I L \sqrt{1 + 1} = \sqrt{2} B_{0} I L$
Hence, magnitude of magnetic force, $| F | = \sqrt{2} B_{0} IL$

MHT-CET 2020

Moving Charges & Magnetism

153758 A metal wire of length ' $L$ ' is bent to form a circular coil of number of turns ' $n$ '. The coil is placed in magnetic field ' $B$ ' and current is passed through the coil. The maximum torque acting on the coil is

1

\frac{B^{2} I L}{2 π n}

2

\frac{{BIL}^{2}}{2 π n}

3

\frac{{BIL}^{2}}{4 π n}

4

\frac{B^{2} I L}{4 π n}

Explanation:

C Given,
$L = 2 π rn$
$r = \frac{L}{2 π n}$
When current I passed through the loop the maximum torque acting on the loop is
$τ = IBAn$
$τ = IB π r^{2} n$
$τ = IB π \frac{L^{2}}{π^{2} n^{2}} \times n$
$τ = IB \frac{L^{2}}{4 π n}$

MHT-CET 2020

Moving Charges & Magnetism

153759 Given figure shows the north and south poles of a permanent magnet in which a coil of $n$ turns of cross-sectional area $A$ is resting, such that when a current $I$ is passed through the coil, the plane of the coil makes and angle $θ$ with respect to direction of magnetic field $B$ . If the plane of magnetic field and the coil are horizontal and vertical respectively, the torque on the coil will be

1

n I A B \cos θ

2

n I A B \sin θ

3 nIAB

4 None of the above, since the magnetic field is radial

Explanation:

A Plane of coil having angle $θ$ with the magnetic field.
$τ = MB \sin (90^{\circ} - θ)$
Magnetic moment $(M) = nIA$
From equation (i) and equation (ii), we get -
$τ = n I A B \cos θ$

AP EAMCET (17.09.2020) Shift-II

Moving Charges & Magnetism

153760 A coil of 50 turns carrying a current of $2 A$ in a magnetic field of $0.5 T$ . The torque acting on the coil is

1

0.4 Nm

clockwise

2

0.2 Nm

anticlockwise

3

0.4 Nm

anticlockwise

4

0.2 Nm

clockwise

5

0.8 Nm

anticlockwise

Explanation:

A Given, Number of turns on the coil, $N = 50$
Current carried by the coil, $I = 2 A$
Magnetic field, $B = 0.5 T$

According to the figure,
The sides $AB$ and $DC$ are along the field lines hence, the force on each side is zero.
The torque on each vertical wire is given as
$τ = B I N A \sin θ$
$τ = 0.5 \times 2 \times 50 \times 0.10 \times 0.08 \sin 90 (∵ \sin θ = 1)$
$τ = 50 \times 0.008$
$τ = 0.4 Nm (clockwise) .$

Kerala CEE 2020

Moving Charges & Magnetism

153755 A wire carrying current ' $I$ ' along $x$ axis has length ' $ℓ$ ' and it is kept in a magnetic field $\vec{B} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) B \frac{W b}{m^{2}}$ . The magnitude of magnetic force acting on the wire is

1

\sqrt{11} I ℓ B

2

\sqrt{15} I ℓ B

3

\sqrt{13} I ℓ B

4

\sqrt{19} I ℓ B

Explanation:

C Given,
$\vec{B} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) B \frac{W b}{m^{2}}$
Current passing along $x$ -axis
$F = i l B$
$F = l (\hat{Ii}) B (\hat{i} + 2 \hat{j} - 3 \hat{k})$
$F = l IB (+ 2 \hat{j} - 3 \hat{k})$
$∣ F = l IB \sqrt{4 + 9}$
$| F | = \sqrt{13} I l B$

MHT-CET 2020

Moving Charges & Magnetism

153757 A wire of length ' $L$ ' carries current ' $I$ ' along $x$ axis, A magnetic field $B = B_{0} (\hat{i} - \hat{j} - \hat{k}) T$ acts on the wire. The magnitude of magnetic force acting on the wire is

1

\frac{{ILB}_{0}}{2}

2 I L B

_{0}

3

\sqrt{2} {ILB}_{0}

4

2 {ILB}_{0}

Explanation:

C Given, $B = B_{0} (\hat{i} - \hat{j} - \hat{k}) T$ and current $= I \hat{i}$
Force acting on the wire
$F = B I L$
$F = L (I \hat{i}) B_{0} (\hat{i} - \hat{j} - \hat{k})$
$F = B_{0} I L (+ \hat{j} - \hat{k})$
$| F | = B_{0} I L \sqrt{1 + 1} = \sqrt{2} B_{0} I L$
Hence, magnitude of magnetic force, $| F | = \sqrt{2} B_{0} IL$

MHT-CET 2020

Moving Charges & Magnetism

153758 A metal wire of length ' $L$ ' is bent to form a circular coil of number of turns ' $n$ '. The coil is placed in magnetic field ' $B$ ' and current is passed through the coil. The maximum torque acting on the coil is

1

\frac{B^{2} I L}{2 π n}

2

\frac{{BIL}^{2}}{2 π n}

3

\frac{{BIL}^{2}}{4 π n}

4

\frac{B^{2} I L}{4 π n}

Explanation:

C Given,
$L = 2 π rn$
$r = \frac{L}{2 π n}$
When current I passed through the loop the maximum torque acting on the loop is
$τ = IBAn$
$τ = IB π r^{2} n$
$τ = IB π \frac{L^{2}}{π^{2} n^{2}} \times n$
$τ = IB \frac{L^{2}}{4 π n}$

MHT-CET 2020

Moving Charges & Magnetism

153759 Given figure shows the north and south poles of a permanent magnet in which a coil of $n$ turns of cross-sectional area $A$ is resting, such that when a current $I$ is passed through the coil, the plane of the coil makes and angle $θ$ with respect to direction of magnetic field $B$ . If the plane of magnetic field and the coil are horizontal and vertical respectively, the torque on the coil will be

1

n I A B \cos θ

2

n I A B \sin θ

3 nIAB

4 None of the above, since the magnetic field is radial

Explanation:

A Plane of coil having angle $θ$ with the magnetic field.
$τ = MB \sin (90^{\circ} - θ)$
Magnetic moment $(M) = nIA$
From equation (i) and equation (ii), we get -
$τ = n I A B \cos θ$

AP EAMCET (17.09.2020) Shift-II

Moving Charges & Magnetism

153760 A coil of 50 turns carrying a current of $2 A$ in a magnetic field of $0.5 T$ . The torque acting on the coil is

1

0.4 Nm

clockwise

2

0.2 Nm

anticlockwise

3

0.4 Nm

anticlockwise

4

0.2 Nm

clockwise

5

0.8 Nm

anticlockwise

Explanation:

A Given, Number of turns on the coil, $N = 50$
Current carried by the coil, $I = 2 A$
Magnetic field, $B = 0.5 T$

According to the figure,
The sides $AB$ and $DC$ are along the field lines hence, the force on each side is zero.
The torque on each vertical wire is given as
$τ = B I N A \sin θ$
$τ = 0.5 \times 2 \times 50 \times 0.10 \times 0.08 \sin 90 (∵ \sin θ = 1)$
$τ = 50 \times 0.008$
$τ = 0.4 Nm (clockwise) .$

Kerala CEE 2020

Moving Charges & Magnetism

153755 A wire carrying current ' $I$ ' along $x$ axis has length ' $ℓ$ ' and it is kept in a magnetic field $\vec{B} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) B \frac{W b}{m^{2}}$ . The magnitude of magnetic force acting on the wire is

1

\sqrt{11} I ℓ B

2

\sqrt{15} I ℓ B

3

\sqrt{13} I ℓ B

4

\sqrt{19} I ℓ B

Explanation:

C Given,
$\vec{B} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) B \frac{W b}{m^{2}}$
Current passing along $x$ -axis
$F = i l B$
$F = l (\hat{Ii}) B (\hat{i} + 2 \hat{j} - 3 \hat{k})$
$F = l IB (+ 2 \hat{j} - 3 \hat{k})$
$∣ F = l IB \sqrt{4 + 9}$
$| F | = \sqrt{13} I l B$

MHT-CET 2020

Moving Charges & Magnetism

153757 A wire of length ' $L$ ' carries current ' $I$ ' along $x$ axis, A magnetic field $B = B_{0} (\hat{i} - \hat{j} - \hat{k}) T$ acts on the wire. The magnitude of magnetic force acting on the wire is

1

\frac{{ILB}_{0}}{2}

2 I L B

_{0}

3

\sqrt{2} {ILB}_{0}

4

2 {ILB}_{0}

Explanation:

C Given, $B = B_{0} (\hat{i} - \hat{j} - \hat{k}) T$ and current $= I \hat{i}$
Force acting on the wire
$F = B I L$
$F = L (I \hat{i}) B_{0} (\hat{i} - \hat{j} - \hat{k})$
$F = B_{0} I L (+ \hat{j} - \hat{k})$
$| F | = B_{0} I L \sqrt{1 + 1} = \sqrt{2} B_{0} I L$
Hence, magnitude of magnetic force, $| F | = \sqrt{2} B_{0} IL$

MHT-CET 2020

Moving Charges & Magnetism

153758 A metal wire of length ' $L$ ' is bent to form a circular coil of number of turns ' $n$ '. The coil is placed in magnetic field ' $B$ ' and current is passed through the coil. The maximum torque acting on the coil is

1

\frac{B^{2} I L}{2 π n}

2

\frac{{BIL}^{2}}{2 π n}

3

\frac{{BIL}^{2}}{4 π n}

4

\frac{B^{2} I L}{4 π n}

Explanation:

C Given,
$L = 2 π rn$
$r = \frac{L}{2 π n}$
When current I passed through the loop the maximum torque acting on the loop is
$τ = IBAn$
$τ = IB π r^{2} n$
$τ = IB π \frac{L^{2}}{π^{2} n^{2}} \times n$
$τ = IB \frac{L^{2}}{4 π n}$

MHT-CET 2020

Moving Charges & Magnetism

153759 Given figure shows the north and south poles of a permanent magnet in which a coil of $n$ turns of cross-sectional area $A$ is resting, such that when a current $I$ is passed through the coil, the plane of the coil makes and angle $θ$ with respect to direction of magnetic field $B$ . If the plane of magnetic field and the coil are horizontal and vertical respectively, the torque on the coil will be

1

n I A B \cos θ

2

n I A B \sin θ

3 nIAB

4 None of the above, since the magnetic field is radial

Explanation:

A Plane of coil having angle $θ$ with the magnetic field.
$τ = MB \sin (90^{\circ} - θ)$
Magnetic moment $(M) = nIA$
From equation (i) and equation (ii), we get -
$τ = n I A B \cos θ$

AP EAMCET (17.09.2020) Shift-II

Moving Charges & Magnetism

153760 A coil of 50 turns carrying a current of $2 A$ in a magnetic field of $0.5 T$ . The torque acting on the coil is

1

0.4 Nm

clockwise

2

0.2 Nm

anticlockwise

3

0.4 Nm

anticlockwise

4

0.2 Nm

clockwise

5

0.8 Nm

anticlockwise

Explanation:

A Given, Number of turns on the coil, $N = 50$
Current carried by the coil, $I = 2 A$
Magnetic field, $B = 0.5 T$

According to the figure,
The sides $AB$ and $DC$ are along the field lines hence, the force on each side is zero.
The torque on each vertical wire is given as
$τ = B I N A \sin θ$
$τ = 0.5 \times 2 \times 50 \times 0.10 \times 0.08 \sin 90 (∵ \sin θ = 1)$
$τ = 50 \times 0.008$
$τ = 0.4 Nm (clockwise) .$

Kerala CEE 2020

Moving Charges & Magnetism

153755 A wire carrying current ' $I$ ' along $x$ axis has length ' $ℓ$ ' and it is kept in a magnetic field $\vec{B} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) B \frac{W b}{m^{2}}$ . The magnitude of magnetic force acting on the wire is

1

\sqrt{11} I ℓ B

2

\sqrt{15} I ℓ B

3

\sqrt{13} I ℓ B

4

\sqrt{19} I ℓ B

Explanation:

C Given,
$\vec{B} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) B \frac{W b}{m^{2}}$
Current passing along $x$ -axis
$F = i l B$
$F = l (\hat{Ii}) B (\hat{i} + 2 \hat{j} - 3 \hat{k})$
$F = l IB (+ 2 \hat{j} - 3 \hat{k})$
$∣ F = l IB \sqrt{4 + 9}$
$| F | = \sqrt{13} I l B$

MHT-CET 2020

Moving Charges & Magnetism

153757 A wire of length ' $L$ ' carries current ' $I$ ' along $x$ axis, A magnetic field $B = B_{0} (\hat{i} - \hat{j} - \hat{k}) T$ acts on the wire. The magnitude of magnetic force acting on the wire is

1

\frac{{ILB}_{0}}{2}

2 I L B

_{0}

3

\sqrt{2} {ILB}_{0}

4

2 {ILB}_{0}

Explanation:

C Given, $B = B_{0} (\hat{i} - \hat{j} - \hat{k}) T$ and current $= I \hat{i}$
Force acting on the wire
$F = B I L$
$F = L (I \hat{i}) B_{0} (\hat{i} - \hat{j} - \hat{k})$
$F = B_{0} I L (+ \hat{j} - \hat{k})$
$| F | = B_{0} I L \sqrt{1 + 1} = \sqrt{2} B_{0} I L$
Hence, magnitude of magnetic force, $| F | = \sqrt{2} B_{0} IL$

MHT-CET 2020

Moving Charges & Magnetism

153758 A metal wire of length ' $L$ ' is bent to form a circular coil of number of turns ' $n$ '. The coil is placed in magnetic field ' $B$ ' and current is passed through the coil. The maximum torque acting on the coil is

1

\frac{B^{2} I L}{2 π n}

2

\frac{{BIL}^{2}}{2 π n}

3

\frac{{BIL}^{2}}{4 π n}

4

\frac{B^{2} I L}{4 π n}

Explanation:

C Given,
$L = 2 π rn$
$r = \frac{L}{2 π n}$
When current I passed through the loop the maximum torque acting on the loop is
$τ = IBAn$
$τ = IB π r^{2} n$
$τ = IB π \frac{L^{2}}{π^{2} n^{2}} \times n$
$τ = IB \frac{L^{2}}{4 π n}$

MHT-CET 2020

Moving Charges & Magnetism

153759 Given figure shows the north and south poles of a permanent magnet in which a coil of $n$ turns of cross-sectional area $A$ is resting, such that when a current $I$ is passed through the coil, the plane of the coil makes and angle $θ$ with respect to direction of magnetic field $B$ . If the plane of magnetic field and the coil are horizontal and vertical respectively, the torque on the coil will be

1

n I A B \cos θ

2

n I A B \sin θ

3 nIAB

4 None of the above, since the magnetic field is radial

Explanation:

A Plane of coil having angle $θ$ with the magnetic field.
$τ = MB \sin (90^{\circ} - θ)$
Magnetic moment $(M) = nIA$
From equation (i) and equation (ii), we get -
$τ = n I A B \cos θ$

AP EAMCET (17.09.2020) Shift-II

Moving Charges & Magnetism

153760 A coil of 50 turns carrying a current of $2 A$ in a magnetic field of $0.5 T$ . The torque acting on the coil is

1

0.4 Nm

clockwise

2

0.2 Nm

anticlockwise

3

0.4 Nm

anticlockwise

4

0.2 Nm

clockwise

5

0.8 Nm

anticlockwise

Explanation:

A Given, Number of turns on the coil, $N = 50$
Current carried by the coil, $I = 2 A$
Magnetic field, $B = 0.5 T$

According to the figure,
The sides $AB$ and $DC$ are along the field lines hence, the force on each side is zero.
The torque on each vertical wire is given as
$τ = B I N A \sin θ$
$τ = 0.5 \times 2 \times 50 \times 0.10 \times 0.08 \sin 90 (∵ \sin θ = 1)$
$τ = 50 \times 0.008$
$τ = 0.4 Nm (clockwise) .$

Kerala CEE 2020

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