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Magnetic Effects of Current and Magnetism MCQs for NEET — Physics Questions with Answers

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Two particles each of mass m and charge q are attached to the two ends of a light rigid rod of length 2R. The rod is rotated at constant angular speed about a perpendicular axis passing through its centre. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod is:

$\frac{q}{2 m}$

$\frac{q}{m}$

$\frac{2 q}{m}$

$\frac{q}{πm}$

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Explanation

(a) $i = \frac{2 q ω}{2 π} = \frac{q ω}{π}; ∴ M = i A = \frac{q ω}{π} {πR}^{2} = q ω R^{2}$

$L = 2 R . m v = 2 R . m R ω = 2 m R^{2} ω (∴ v = R ω)$

$\Rightarrow \frac{M}{L} = \frac{q}{2 m}$

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If m is the magnetic moment and B is the magnetic field, then the torque is given by

$\vec{m} . \vec{B}$

$\frac{| \vec{m} |}{| \vec{B} |}$

$\vec{m} \times \vec{B}$

$| \vec{m} | . | \vec{B} |$

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Explanation

(c)

$T o r q u e = \vec{m} \times \vec{B}$

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A 250-turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of 85 $μ$ A and subjected to a magnetic field of strength o.85 T. Work done for rotating the coil by $180^{\circ}$ against the torque is

9.1 $μ J$

4.55 $μ J$

2.3 $μ$ $J$

1.5 $μ J$

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Explanation

(a) Work done for rotating the coil

W=MB( $\cos θ_{1} - \cos θ_{2}$ )

Where, M=maganetic moment

B=maganetic field

Given. $θ_{1} = O^{\circ}, θ_{2} = 180^{\circ}$

$\Rightarrow$ W=MB( $(\cos 0^{°} - \cos 180^{°})$

= 2MB=2 $\times$ NIA $\times B$

= $2 \times 250 \times 85 \times 10^{- 6} {(1.25 \times 2.1 \times 10^{- 4})}_{} \times 85 \times 10^{- 2}$

=9.1 $μ J$ (Approx)

The closest option is (a).

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A current loop in a magnetic field

experiences a torque whether the field is uniform or non-uniform in all orientations

can be in equilibrium in one orentations.

can be equilibrium in two orientations, both the wquilibriu states are unstable

can be in equilibrium in two orientations, one stable while the other is unstable

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Explanation

(d) For parallel M is stable and for antiparallel is unstable.

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A closely wound solenoid of 2000 turns and area of cross-section $1.5 \times 10^{- 4} m^{2}$ carries a current of $2.0 A .$ It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5 \times 10^{- 2} T$ making an angle of $30 °$ with the axis of the solenoid. The torque on the solenoid will be

$3 \times 10^{- 3} N ‐ m$

$1.5 \times 10^{- 3} N ‐ m$

$1.5 \times 10^{- 2} N ‐ m$

$3 \times 10^{- 2} N ‐ m$

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Explanation

$Given, N = 2000, A = 1.5 \times 10^{- 4} m^{2} i = 2.0 A B = 5 \times 10^{- 2} T, and θ = 30 ° Torque, τ = N i B A \sin θ = 2000 \times 2 \times 5 \times 10^{- 2} \times 1.5 \times 10^{- 4} \times \sin 30 ° = 2000 \times 50 \times 10^{- 6} \times \frac{1}{2} = 1.5 \times 10^{- 2} Nm$

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A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in the equilibrium state. The energy required to rotate it by 60^o is W. Now the torque required to keep the magnet in this new position is:

$\frac{W}{\sqrt{3}}$

$\sqrt{3} W$

$\frac{\sqrt{3} W}{2}$

$\frac{2 W}{\sqrt{3}}$

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Explanation

Torque = MBsinθ
= MB sin $60^{0}$ -------(1)
Work done in displacing the magnet from an angle $θ_{1}$ to $θ_{2}$ is -
W = MB(cos $θ_{1}$ – cos $θ_{2}$ )
W = MB(1 – cos $60^{0}$ ) -------(2)
From (1) and (2)
Torque= $\frac{\frac{\sqrt{3} W}{2}}{\frac{1}{} 2} = \sqrt{3} W$

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A 250-Turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of 85 $μ$ and subjected to the magnetic field of strength 0.85 T. Work done for rotating the coil by 180° against the torque is:

$4.55 μ J$

$2.3 μ J$

$1.15 μ J$

$9.8 μ J$

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Explanation

Given a rectangular coil of length 2.1 cm and width 1.25 cm
Current through coil = 85 $μ A$
No. of turns = 250
B= 0.85 T
Work done, W= MB $(\cos θ_{1} - \cos θ_{2})$
When it is rotated by angle 180⁰ then
W= MB $(\cos 0^{0} - \cos 180^{0}) = M B (1 + 1) = 2 M B$
W = 2(NIA)B
W= $2 \times 250 \times 85 \times 10^{- 6} [1.25 \times 2.1 \times 10^{- 4}] \times 85 \times 10^{- 2}$

W= $9.8 μ J$

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A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Wb/m². The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of 30^o with the direction of the field, the torque required to keep the coil in stable equilibrium will be:

0.15 Nm

0.20 Nm

0.24 Nm

0.12 Nm

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A bar magnet of length l and magnetic dipole moment M is bent to form an arc which subtends an angle of $120 °$ at centre. The new magnetic dipole moment will be

$\frac{3 M}{2 π}$

$\frac{3 \sqrt{3} M}{2 π}$

$\frac{3 M}{π}$

$\frac{2 M}{π}$

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Explanation

When a bar magnet is bent into an arc, its magnetic dipole moment changes. The new magnetic dipole moment is directly proportional to the original dipole moment and the angle subtended by the arc at the center. For an angle of 120Â°, the new dipole moment is (3âˆš3/2Ï€)M.

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The unit of pole strength is

$A m^{2}$

$\frac{A^{2}}{m}$

$\frac{A^{2}}{m^{2}}$

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Explanation

If we define magnetic moment like electric dipole moment

Magnetic moment ( Ampere X $m^{2}$ ) = Pole strength X Distance (m)

Unit of pole strength = Ampere X meter

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