Magnetic Effects of Current

Practice Questions:

Question 11: moderate

A charge Q is uniformly distributed over the surface of non-conducting disc of radius R. The
disc rotates about an axis perpendicular to its plane and passing through its centre with an
angular velocity ω. As a result of this rotation a magnetic field of induction B is obtained at
the centre of the disc. If we keep both the amount of charge placed on the disc and its
angular velocity to be constant and vary the radius of the disc then the variation of the
magnetic induction at the centre of the disc will be represented by the figure

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Question 12: moderate

A current I is flowing through the loop. The direction of the current and the shape of the loop are as shown in the figure. The magnetic field at the centre of the loop is (MA = R, MB = 2R, angle DMA = 90°)

1. \[\frac{7\mu_{0}i}{16R}\] , but out of the plane of the paper.

2. \[\frac{5\mu_{0}i}{16R}\] , but out of the plane of the paper.

3. \[\frac{7\mu_{0}i}{16R}\] , but into the plane of the paper.

4. \[\frac{5\mu_{0}i}{16R}\] , but into the plane of the paper.

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Question 13: moderate

In the adjacent figure is shown a closed path P. A long straight conductor carrying a current I passes through O and perpendicular to the plane of the paper. Then which of the following holds good ?

1. \[\int_{P}^{}\overrightarrow{B}.\overrightarrow{dl}=0\]

2. \[\int_{P}^{}\overrightarrow{B}.\overrightarrow{dl}=\mu_{0}I\]

3. \[\int_{P}^{}\overrightarrow{B}.\overrightarrow{dl}>\mu_{0}I\]

4. None of these

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As Current enclosed is zero. \[\int_{P}^{}\overrightarrow{B}.\overrightarrow{dl}=0\]

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Question 14: moderate

A bar magnet of length ‘l’ and magnetic dipole moment ‘M’ is bent in the form of an arc as
shown in figure. The new magnetic dipole moment will be :

1. M/2

2. M

3. \[\frac{3}{\pi}M\]

4. \[\frac{2}{\pi}M\]

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Question 15: moderate

A loop carrying current I lies in the x–y plane as shown in the figure. The unit vector ˆk is coming out of the plane of the paper. The magnetic moment of the current loop is :

1. \[a^{2}I\hat{k}\]

2. \[\left( \frac{\pi}{2}+1 \right)a^{2}I\hat{k}\]

3. \[-\left( \frac{\pi}{2}+1 \right)a^{2}I\hat{k}\]

4. \[\left( 2\pi+1 \right)a^{2}I\hat{k}\]

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Question 16: moderate

A coil having N turns is would tightly in the form of a spiral with inner and outer radii a and b respectively. When a current I passes through the coil, the magnetic field at its centre is :

1. \[\frac{\mu_{0}NI}{b}\]

2. \[\frac{2\mu_{0}NI}{a}\]

3. \[\frac{\mu_{0}NI}{2\left( b-a \right)}log\frac{b}{a}\]

4. \[\frac{\mu_{0}NI}{\left( b-a \right)}log\frac{b}{a}\]

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Question 17: moderate

A particle carrying a charge equal to 100 times the charge on an electron is rotating per second in a circular path of radius 0.8m. The value of the magnetic field produced at the centre will be : (μ0 = permeability constant)

1. \[10^{-7}/\mu_{0}\]

2. \[10^{-17}\mu_{0}\]

3. \[10^{-6}/\mu_{0}\]

4. \[10^{-7}\mu_{0}\]

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Question 18: moderate

The magnetic induction at the centre O of the current carrying bent wire shown in the adjoining figure is :

1. \[\frac{\mu_{0}I}{4\pi R_{1}}\alpha\]

2. \[\frac{\mu_{0}I}{4\pi R_{2}}\alpha\]

3. \[\frac{\mu_{0}I\alpha}{4\pi}\left( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right)\]

4. \[\frac{\mu_{0}I\alpha}{4\pi}\left( \frac{1}{R_{1}}+\frac{1}{R_{2}} \right)\]

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Question 19: moderate

A cell is connected between the points A and C of a circular conductor ABCD with O as centre and angle AOC = 60°. If B1 and B2 are the magnitudes of the magnetic fields at O due to the currents in ABC and ADC respectively, then ratio B1/B2 is :-

1. 1

2. 2

3. 5

4. 6

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Question 20: moderate

Consider six wires coming into or out of the page, all with the same current. Rank the line integral of the magnetic field (from most positive to most negative) taken counter-clockwise around each

loop shown.

1. B > C > D > A

2. B > C = D > A

3. B > A > C = D

4. C > B = D > A

View Answer

Current enclosed in A = 3i -3i =0

Current enclosed in B = 2i-i=i

Current enclosed in C = - 2i + i= - i

Current enclosed in D= -i

so Correct order of integral of B.dl is B > A > C = D

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