QBManager

Moving Charges & Magnetism

153539 A particle with charge $1.6 \times 10^{- 5} C$ is moving with a velocity of magnitude $3 m / s$ in uniform magnetic field of $2 T$ . The direction of velocity remains perpendicular to the direction of magnetic field. What is the magnitude of net force on the particle?

1

4.6 \times 10^{- 5} N

2

9.6 \times 10^{- 5} N

3

5.6 \times 10^{- 5} N

4

12.2 \times 10^{- 5} N

Explanation:

B Given,
$q = 1.6 \times 10^{- 5} C$
$v = 3 m / \sec$
$B = 2 T$
$θ = 90^{\circ}$
Magnitude of net force particle
$F = q v vB \sin θ$
$F = 1.6 \times 10^{- 5} \times 3 \times 2 \times \sin 90^{\circ}$
$F = 9.6 \times 10^{- 5} N$

TS EAMCET 09.05.2019

Moving Charges & Magnetism

153540 A proton moving in perpendicular magnetic field possesses energy ' $E$ '. The magnetic field is increased four times. But the proton is constrained to move in the path of same radius. The kinetic energy will increase

1 2 times

2 16 times

3 8 times

4 4 times

Explanation:

B Force due to magnetic field.
$F = q (v \times B)$
We know that,
$\frac{{mv}^{2}}{R} = qvB$
$v = \frac{qBR}{m}$
Kinetic energy (K.E.) $= (\frac{1}{2}) {mv}^{2}$
Putting the value of ' $v$ ' in equation (i), we get -
$K.E. = \frac{p^{2}}{2 m} = \frac{q^{2} B^{2} R^{2}}{2 m}$
Charge (q), mass (m) and radius (R) are constant -
$K.E. = B^{2}$
$K.E. = (4)^{2}$
$K.E. = 16$
Hence, the kinetic energy increases 16 times.

MHT-CET 2019

Moving Charges & Magnetism

153541 The work done by an uniform magnetic field, on a moving charge is

1 zero because

\vec{F}

acts parallel to

\vec{V}

2 positive because

\vec{F}

acts perpendicular to

\vec{V}

3 zero because

\vec{F}

acts perpendicular to

\vec{V}

4 negative because

\vec{F}

acts parallel to

\vec{v}

Explanation:

C Force on moving charge while moving in magnetic field
$\vec{F} = q (\vec{v} \times \vec{B})$
Where, $q =$ charge of the particle
$v = velocity of particle$
$B = magnetic field$
Where $\vec{F}$ is perpendicular to $\vec{V}$ .
Work done/sec,
$W = \vec{F} \cdot \vec{v}$
$W = \vec{F} \cdot \vec{v} \cos θ$
Since they are perpendicular, then $θ = 90^{\circ}$
$W = F \cdot v \cos 90^{\circ}$
$W = 0$
Hence, the work done by an uniform magnetic field, on a moving charge is zero because $\vec{F}$ acts perpendicular to $\vec{v}$ .

VITEEE-2019

Moving Charges & Magnetism

153542 Assertion: A charge particle is released from rest in magnetic field then it will move in circular path.
Reason: Work done by magnetic field is non zero.

1 If both assertion and reason are true and reason is the correct explanation of assertion.

2 If both assertion and reason are true but reason is not the correct explanation of assertion.

3 If assertion is true but reason is false.

4 If both assertion and reason are false.

Explanation:

D Force due to electric field $(F_{e}) = q \cdot E$
Where, $q =$ charge
Force due to magnetic field,
$| F_{m} | = q (\vec{v} \times \vec{B}) (∵ a \times b = ab \sin θ)$
$| F_{m} | = q \cdot v \cdot B \sin θ$
$F_{m} = q \cdot v \cdot B \sin θ$
$F_{m} = 0$
A charged particle is released from rest in magnetic field then it will not move in circular path because work done by magnetic field is zero.
Hence, both assertion and reason are false.

AIIMS-26.05.2019(E) Shift-2

Moving Charges & Magnetism

153544 The magnetic force acting on a charged particle of charge $- 2 μ C$ in a magnetic field of 2 $T$ acting in $y$ direction, when the particle velocity is $(2 \hat{i} + 3 \hat{j}) \times 10^{6} m s^{- 1}$ , is

1

4 N

in

z

- direction

2

8 N

in

y

- direction

3

8 N

in

x

- direction

4

8 N

in

z

- direction

Explanation:

D Given that,
$q = - 2 μ C = - 2 \times 10^{- 6} C$
$B = 2 T = 2 \hat{j} T$
$v = (2 \hat{i} + 3 \hat{j}) \times 10^{6} m / \sec$
We know that,
Force on moving charge in magnetic field,
$F = q (\vec{v} \times \vec{B})$
$F = - 2 \times 10^{- 6} [(2 \hat{i} + 3 \hat{j}) \times 2 \hat{j}] \times 10^{6}$
$F = - 2 [2 \hat{i} \times 2 \hat{j}] (∵ \hat{i} \times \hat{j} = \hat{k})$
$\vec{F} = - 8 \hat{k} N$
Hence, magnetic force is $8 N$ in $z$ - direction

COMEDK 2019

Moving Charges & Magnetism

153539 A particle with charge $1.6 \times 10^{- 5} C$ is moving with a velocity of magnitude $3 m / s$ in uniform magnetic field of $2 T$ . The direction of velocity remains perpendicular to the direction of magnetic field. What is the magnitude of net force on the particle?

1

4.6 \times 10^{- 5} N

2

9.6 \times 10^{- 5} N

3

5.6 \times 10^{- 5} N

4

12.2 \times 10^{- 5} N

Explanation:

B Given,
$q = 1.6 \times 10^{- 5} C$
$v = 3 m / \sec$
$B = 2 T$
$θ = 90^{\circ}$
Magnitude of net force particle
$F = q v vB \sin θ$
$F = 1.6 \times 10^{- 5} \times 3 \times 2 \times \sin 90^{\circ}$
$F = 9.6 \times 10^{- 5} N$

TS EAMCET 09.05.2019

Moving Charges & Magnetism

153540 A proton moving in perpendicular magnetic field possesses energy ' $E$ '. The magnetic field is increased four times. But the proton is constrained to move in the path of same radius. The kinetic energy will increase

1 2 times

2 16 times

3 8 times

4 4 times

Explanation:

B Force due to magnetic field.
$F = q (v \times B)$
We know that,
$\frac{{mv}^{2}}{R} = qvB$
$v = \frac{qBR}{m}$
Kinetic energy (K.E.) $= (\frac{1}{2}) {mv}^{2}$
Putting the value of ' $v$ ' in equation (i), we get -
$K.E. = \frac{p^{2}}{2 m} = \frac{q^{2} B^{2} R^{2}}{2 m}$
Charge (q), mass (m) and radius (R) are constant -
$K.E. = B^{2}$
$K.E. = (4)^{2}$
$K.E. = 16$
Hence, the kinetic energy increases 16 times.

MHT-CET 2019

Moving Charges & Magnetism

153541 The work done by an uniform magnetic field, on a moving charge is

1 zero because

\vec{F}

acts parallel to

\vec{V}

2 positive because

\vec{F}

acts perpendicular to

\vec{V}

3 zero because

\vec{F}

acts perpendicular to

\vec{V}

4 negative because

\vec{F}

acts parallel to

\vec{v}

Explanation:

C Force on moving charge while moving in magnetic field
$\vec{F} = q (\vec{v} \times \vec{B})$
Where, $q =$ charge of the particle
$v = velocity of particle$
$B = magnetic field$
Where $\vec{F}$ is perpendicular to $\vec{V}$ .
Work done/sec,
$W = \vec{F} \cdot \vec{v}$
$W = \vec{F} \cdot \vec{v} \cos θ$
Since they are perpendicular, then $θ = 90^{\circ}$
$W = F \cdot v \cos 90^{\circ}$
$W = 0$
Hence, the work done by an uniform magnetic field, on a moving charge is zero because $\vec{F}$ acts perpendicular to $\vec{v}$ .

VITEEE-2019

Moving Charges & Magnetism

153542 Assertion: A charge particle is released from rest in magnetic field then it will move in circular path.
Reason: Work done by magnetic field is non zero.

1 If both assertion and reason are true and reason is the correct explanation of assertion.

2 If both assertion and reason are true but reason is not the correct explanation of assertion.

3 If assertion is true but reason is false.

4 If both assertion and reason are false.

Explanation:

D Force due to electric field $(F_{e}) = q \cdot E$
Where, $q =$ charge
Force due to magnetic field,
$| F_{m} | = q (\vec{v} \times \vec{B}) (∵ a \times b = ab \sin θ)$
$| F_{m} | = q \cdot v \cdot B \sin θ$
$F_{m} = q \cdot v \cdot B \sin θ$
$F_{m} = 0$
A charged particle is released from rest in magnetic field then it will not move in circular path because work done by magnetic field is zero.
Hence, both assertion and reason are false.

AIIMS-26.05.2019(E) Shift-2

Moving Charges & Magnetism

153544 The magnetic force acting on a charged particle of charge $- 2 μ C$ in a magnetic field of 2 $T$ acting in $y$ direction, when the particle velocity is $(2 \hat{i} + 3 \hat{j}) \times 10^{6} m s^{- 1}$ , is

1

4 N

in

z

- direction

2

8 N

in

y

- direction

3

8 N

in

x

- direction

4

8 N

in

z

- direction

Explanation:

D Given that,
$q = - 2 μ C = - 2 \times 10^{- 6} C$
$B = 2 T = 2 \hat{j} T$
$v = (2 \hat{i} + 3 \hat{j}) \times 10^{6} m / \sec$
We know that,
Force on moving charge in magnetic field,
$F = q (\vec{v} \times \vec{B})$
$F = - 2 \times 10^{- 6} [(2 \hat{i} + 3 \hat{j}) \times 2 \hat{j}] \times 10^{6}$
$F = - 2 [2 \hat{i} \times 2 \hat{j}] (∵ \hat{i} \times \hat{j} = \hat{k})$
$\vec{F} = - 8 \hat{k} N$
Hence, magnetic force is $8 N$ in $z$ - direction

COMEDK 2019

Moving Charges & Magnetism

153539 A particle with charge $1.6 \times 10^{- 5} C$ is moving with a velocity of magnitude $3 m / s$ in uniform magnetic field of $2 T$ . The direction of velocity remains perpendicular to the direction of magnetic field. What is the magnitude of net force on the particle?

1

4.6 \times 10^{- 5} N

2

9.6 \times 10^{- 5} N

3

5.6 \times 10^{- 5} N

4

12.2 \times 10^{- 5} N

Explanation:

B Given,
$q = 1.6 \times 10^{- 5} C$
$v = 3 m / \sec$
$B = 2 T$
$θ = 90^{\circ}$
Magnitude of net force particle
$F = q v vB \sin θ$
$F = 1.6 \times 10^{- 5} \times 3 \times 2 \times \sin 90^{\circ}$
$F = 9.6 \times 10^{- 5} N$

TS EAMCET 09.05.2019

Moving Charges & Magnetism

153540 A proton moving in perpendicular magnetic field possesses energy ' $E$ '. The magnetic field is increased four times. But the proton is constrained to move in the path of same radius. The kinetic energy will increase

1 2 times

2 16 times

3 8 times

4 4 times

Explanation:

B Force due to magnetic field.
$F = q (v \times B)$
We know that,
$\frac{{mv}^{2}}{R} = qvB$
$v = \frac{qBR}{m}$
Kinetic energy (K.E.) $= (\frac{1}{2}) {mv}^{2}$
Putting the value of ' $v$ ' in equation (i), we get -
$K.E. = \frac{p^{2}}{2 m} = \frac{q^{2} B^{2} R^{2}}{2 m}$
Charge (q), mass (m) and radius (R) are constant -
$K.E. = B^{2}$
$K.E. = (4)^{2}$
$K.E. = 16$
Hence, the kinetic energy increases 16 times.

MHT-CET 2019

Moving Charges & Magnetism

153541 The work done by an uniform magnetic field, on a moving charge is

1 zero because

\vec{F}

acts parallel to

\vec{V}

2 positive because

\vec{F}

acts perpendicular to

\vec{V}

3 zero because

\vec{F}

acts perpendicular to

\vec{V}

4 negative because

\vec{F}

acts parallel to

\vec{v}

Explanation:

C Force on moving charge while moving in magnetic field
$\vec{F} = q (\vec{v} \times \vec{B})$
Where, $q =$ charge of the particle
$v = velocity of particle$
$B = magnetic field$
Where $\vec{F}$ is perpendicular to $\vec{V}$ .
Work done/sec,
$W = \vec{F} \cdot \vec{v}$
$W = \vec{F} \cdot \vec{v} \cos θ$
Since they are perpendicular, then $θ = 90^{\circ}$
$W = F \cdot v \cos 90^{\circ}$
$W = 0$
Hence, the work done by an uniform magnetic field, on a moving charge is zero because $\vec{F}$ acts perpendicular to $\vec{v}$ .

VITEEE-2019

Moving Charges & Magnetism

153542 Assertion: A charge particle is released from rest in magnetic field then it will move in circular path.
Reason: Work done by magnetic field is non zero.

1 If both assertion and reason are true and reason is the correct explanation of assertion.

2 If both assertion and reason are true but reason is not the correct explanation of assertion.

3 If assertion is true but reason is false.

4 If both assertion and reason are false.

Explanation:

D Force due to electric field $(F_{e}) = q \cdot E$
Where, $q =$ charge
Force due to magnetic field,
$| F_{m} | = q (\vec{v} \times \vec{B}) (∵ a \times b = ab \sin θ)$
$| F_{m} | = q \cdot v \cdot B \sin θ$
$F_{m} = q \cdot v \cdot B \sin θ$
$F_{m} = 0$
A charged particle is released from rest in magnetic field then it will not move in circular path because work done by magnetic field is zero.
Hence, both assertion and reason are false.

AIIMS-26.05.2019(E) Shift-2

Moving Charges & Magnetism

153544 The magnetic force acting on a charged particle of charge $- 2 μ C$ in a magnetic field of 2 $T$ acting in $y$ direction, when the particle velocity is $(2 \hat{i} + 3 \hat{j}) \times 10^{6} m s^{- 1}$ , is

1

4 N

in

z

- direction

2

8 N

in

y

- direction

3

8 N

in

x

- direction

4

8 N

in

z

- direction

Explanation:

D Given that,
$q = - 2 μ C = - 2 \times 10^{- 6} C$
$B = 2 T = 2 \hat{j} T$
$v = (2 \hat{i} + 3 \hat{j}) \times 10^{6} m / \sec$
We know that,
Force on moving charge in magnetic field,
$F = q (\vec{v} \times \vec{B})$
$F = - 2 \times 10^{- 6} [(2 \hat{i} + 3 \hat{j}) \times 2 \hat{j}] \times 10^{6}$
$F = - 2 [2 \hat{i} \times 2 \hat{j}] (∵ \hat{i} \times \hat{j} = \hat{k})$
$\vec{F} = - 8 \hat{k} N$
Hence, magnetic force is $8 N$ in $z$ - direction

COMEDK 2019

Moving Charges & Magnetism

153539 A particle with charge $1.6 \times 10^{- 5} C$ is moving with a velocity of magnitude $3 m / s$ in uniform magnetic field of $2 T$ . The direction of velocity remains perpendicular to the direction of magnetic field. What is the magnitude of net force on the particle?

1

4.6 \times 10^{- 5} N

2

9.6 \times 10^{- 5} N

3

5.6 \times 10^{- 5} N

4

12.2 \times 10^{- 5} N

Explanation:

B Given,
$q = 1.6 \times 10^{- 5} C$
$v = 3 m / \sec$
$B = 2 T$
$θ = 90^{\circ}$
Magnitude of net force particle
$F = q v vB \sin θ$
$F = 1.6 \times 10^{- 5} \times 3 \times 2 \times \sin 90^{\circ}$
$F = 9.6 \times 10^{- 5} N$

TS EAMCET 09.05.2019

Moving Charges & Magnetism

153540 A proton moving in perpendicular magnetic field possesses energy ' $E$ '. The magnetic field is increased four times. But the proton is constrained to move in the path of same radius. The kinetic energy will increase

1 2 times

2 16 times

3 8 times

4 4 times

Explanation:

B Force due to magnetic field.
$F = q (v \times B)$
We know that,
$\frac{{mv}^{2}}{R} = qvB$
$v = \frac{qBR}{m}$
Kinetic energy (K.E.) $= (\frac{1}{2}) {mv}^{2}$
Putting the value of ' $v$ ' in equation (i), we get -
$K.E. = \frac{p^{2}}{2 m} = \frac{q^{2} B^{2} R^{2}}{2 m}$
Charge (q), mass (m) and radius (R) are constant -
$K.E. = B^{2}$
$K.E. = (4)^{2}$
$K.E. = 16$
Hence, the kinetic energy increases 16 times.

MHT-CET 2019

Moving Charges & Magnetism

153541 The work done by an uniform magnetic field, on a moving charge is

1 zero because

\vec{F}

acts parallel to

\vec{V}

2 positive because

\vec{F}

acts perpendicular to

\vec{V}

3 zero because

\vec{F}

acts perpendicular to

\vec{V}

4 negative because

\vec{F}

acts parallel to

\vec{v}

Explanation:

C Force on moving charge while moving in magnetic field
$\vec{F} = q (\vec{v} \times \vec{B})$
Where, $q =$ charge of the particle
$v = velocity of particle$
$B = magnetic field$
Where $\vec{F}$ is perpendicular to $\vec{V}$ .
Work done/sec,
$W = \vec{F} \cdot \vec{v}$
$W = \vec{F} \cdot \vec{v} \cos θ$
Since they are perpendicular, then $θ = 90^{\circ}$
$W = F \cdot v \cos 90^{\circ}$
$W = 0$
Hence, the work done by an uniform magnetic field, on a moving charge is zero because $\vec{F}$ acts perpendicular to $\vec{v}$ .

VITEEE-2019

Moving Charges & Magnetism

153542 Assertion: A charge particle is released from rest in magnetic field then it will move in circular path.
Reason: Work done by magnetic field is non zero.

1 If both assertion and reason are true and reason is the correct explanation of assertion.

2 If both assertion and reason are true but reason is not the correct explanation of assertion.

3 If assertion is true but reason is false.

4 If both assertion and reason are false.

Explanation:

D Force due to electric field $(F_{e}) = q \cdot E$
Where, $q =$ charge
Force due to magnetic field,
$| F_{m} | = q (\vec{v} \times \vec{B}) (∵ a \times b = ab \sin θ)$
$| F_{m} | = q \cdot v \cdot B \sin θ$
$F_{m} = q \cdot v \cdot B \sin θ$
$F_{m} = 0$
A charged particle is released from rest in magnetic field then it will not move in circular path because work done by magnetic field is zero.
Hence, both assertion and reason are false.

AIIMS-26.05.2019(E) Shift-2

Moving Charges & Magnetism

153544 The magnetic force acting on a charged particle of charge $- 2 μ C$ in a magnetic field of 2 $T$ acting in $y$ direction, when the particle velocity is $(2 \hat{i} + 3 \hat{j}) \times 10^{6} m s^{- 1}$ , is

1

4 N

in

z

- direction

2

8 N

in

y

- direction

3

8 N

in

x

- direction

4

8 N

in

z

- direction

Explanation:

D Given that,
$q = - 2 μ C = - 2 \times 10^{- 6} C$
$B = 2 T = 2 \hat{j} T$
$v = (2 \hat{i} + 3 \hat{j}) \times 10^{6} m / \sec$
We know that,
Force on moving charge in magnetic field,
$F = q (\vec{v} \times \vec{B})$
$F = - 2 \times 10^{- 6} [(2 \hat{i} + 3 \hat{j}) \times 2 \hat{j}] \times 10^{6}$
$F = - 2 [2 \hat{i} \times 2 \hat{j}] (∵ \hat{i} \times \hat{j} = \hat{k})$
$\vec{F} = - 8 \hat{k} N$
Hence, magnetic force is $8 N$ in $z$ - direction

COMEDK 2019

Moving Charges & Magnetism

153539 A particle with charge $1.6 \times 10^{- 5} C$ is moving with a velocity of magnitude $3 m / s$ in uniform magnetic field of $2 T$ . The direction of velocity remains perpendicular to the direction of magnetic field. What is the magnitude of net force on the particle?

1

4.6 \times 10^{- 5} N

2

9.6 \times 10^{- 5} N

3

5.6 \times 10^{- 5} N

4

12.2 \times 10^{- 5} N

Explanation:

B Given,
$q = 1.6 \times 10^{- 5} C$
$v = 3 m / \sec$
$B = 2 T$
$θ = 90^{\circ}$
Magnitude of net force particle
$F = q v vB \sin θ$
$F = 1.6 \times 10^{- 5} \times 3 \times 2 \times \sin 90^{\circ}$
$F = 9.6 \times 10^{- 5} N$

TS EAMCET 09.05.2019

Moving Charges & Magnetism

153540 A proton moving in perpendicular magnetic field possesses energy ' $E$ '. The magnetic field is increased four times. But the proton is constrained to move in the path of same radius. The kinetic energy will increase

1 2 times

2 16 times

3 8 times

4 4 times

Explanation:

B Force due to magnetic field.
$F = q (v \times B)$
We know that,
$\frac{{mv}^{2}}{R} = qvB$
$v = \frac{qBR}{m}$
Kinetic energy (K.E.) $= (\frac{1}{2}) {mv}^{2}$
Putting the value of ' $v$ ' in equation (i), we get -
$K.E. = \frac{p^{2}}{2 m} = \frac{q^{2} B^{2} R^{2}}{2 m}$
Charge (q), mass (m) and radius (R) are constant -
$K.E. = B^{2}$
$K.E. = (4)^{2}$
$K.E. = 16$
Hence, the kinetic energy increases 16 times.

MHT-CET 2019

Moving Charges & Magnetism

153541 The work done by an uniform magnetic field, on a moving charge is

1 zero because

\vec{F}

acts parallel to

\vec{V}

2 positive because

\vec{F}

acts perpendicular to

\vec{V}

3 zero because

\vec{F}

acts perpendicular to

\vec{V}

4 negative because

\vec{F}

acts parallel to

\vec{v}

Explanation:

C Force on moving charge while moving in magnetic field
$\vec{F} = q (\vec{v} \times \vec{B})$
Where, $q =$ charge of the particle
$v = velocity of particle$
$B = magnetic field$
Where $\vec{F}$ is perpendicular to $\vec{V}$ .
Work done/sec,
$W = \vec{F} \cdot \vec{v}$
$W = \vec{F} \cdot \vec{v} \cos θ$
Since they are perpendicular, then $θ = 90^{\circ}$
$W = F \cdot v \cos 90^{\circ}$
$W = 0$
Hence, the work done by an uniform magnetic field, on a moving charge is zero because $\vec{F}$ acts perpendicular to $\vec{v}$ .

VITEEE-2019

Moving Charges & Magnetism

153542 Assertion: A charge particle is released from rest in magnetic field then it will move in circular path.
Reason: Work done by magnetic field is non zero.

1 If both assertion and reason are true and reason is the correct explanation of assertion.

2 If both assertion and reason are true but reason is not the correct explanation of assertion.

3 If assertion is true but reason is false.

4 If both assertion and reason are false.

Explanation:

D Force due to electric field $(F_{e}) = q \cdot E$
Where, $q =$ charge
Force due to magnetic field,
$| F_{m} | = q (\vec{v} \times \vec{B}) (∵ a \times b = ab \sin θ)$
$| F_{m} | = q \cdot v \cdot B \sin θ$
$F_{m} = q \cdot v \cdot B \sin θ$
$F_{m} = 0$
A charged particle is released from rest in magnetic field then it will not move in circular path because work done by magnetic field is zero.
Hence, both assertion and reason are false.

AIIMS-26.05.2019(E) Shift-2

Moving Charges & Magnetism

153544 The magnetic force acting on a charged particle of charge $- 2 μ C$ in a magnetic field of 2 $T$ acting in $y$ direction, when the particle velocity is $(2 \hat{i} + 3 \hat{j}) \times 10^{6} m s^{- 1}$ , is

1

4 N

in

z

- direction

2

8 N

in

y

- direction

3

8 N

in

x

- direction

4

8 N

in

z

- direction

Explanation:

D Given that,
$q = - 2 μ C = - 2 \times 10^{- 6} C$
$B = 2 T = 2 \hat{j} T$
$v = (2 \hat{i} + 3 \hat{j}) \times 10^{6} m / \sec$
We know that,
Force on moving charge in magnetic field,
$F = q (\vec{v} \times \vec{B})$
$F = - 2 \times 10^{- 6} [(2 \hat{i} + 3 \hat{j}) \times 2 \hat{j}] \times 10^{6}$
$F = - 2 [2 \hat{i} \times 2 \hat{j}] (∵ \hat{i} \times \hat{j} = \hat{k})$
$\vec{F} = - 8 \hat{k} N$
Hence, magnetic force is $8 N$ in $z$ - direction

COMEDK 2019

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