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This is one of our many quizzes on Current and magnetism. Click on this to access more such competitive quizzes on Electrostatics and Magnetism.

Current and magnetism Quiz 2

physics

 Due to which factor, Maxwell had to modify Ampere’s circuital law ?   

 Displacement current is defined as the rate of change of electric displacement field and its unit is the same as that of electric current density. This concept was introduced to make the Ampere circuit law logically consistent. 

  Maxwell came up with the idea of displacement current which is the electric field derivative term. He added the displacement current to the real current term in Ampere's law, indicating that just as changing magnetic fields give rise to electric fields, changing electric fields give rise to magnetic fields.

 Consider the given condition regarding two wires conduction electric current through them.Then, $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/86789080-4694-4dc7-a153-37a726f0a2798945030103647285164.png"/>

 By using Fleming's rule, find the direction of the magnetic field due to both wires and then check the direction of both forces whether they seem attractive or repulsive. 

  The direction of the magnetic field due to a single wire can be found using Fleming's right hand rule. Now, while considering the middle portion of the two conducting wires, two forces are found to be attractive to each other. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d9f07541-30a7-4586-a183-72b9f0a91483548489873135332407.png"/>

 Which of the given statements hold true for the diagram given? $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4c1658f3-bb5d-4a33-8cc0-4003c7d19eaa7843529547842027354.png"/>

 The direction of emf induced in a coil is given by Lenz’s law. Current induced will create a magnetic field which opposes the buildup of magnetic field in the coil. 

 According to Faraday’s law, if a coil is moving into a magnetic field, there will be a difference in magnetic flux and e.m.f will be generated. The current induced will create a magnetic field which opposes the buildup of magnetic field in the coil.Here, the induced current will be developed in anti-clockwise direction to oppose the build up of magnetic field in the coil. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/c3c6bcb5-183f-45c6-bde4-e77dd25ce7216607016996003385726.png"/>

 If you bring a charge inside the center of a toroid, then

  By using Ampere’s circuital law, find the measure of magnetic field at the center of a toroid and hence we can get to know whether it can have any effect on the charge placed there. 

 Let $B$ be the magnetic field inside the toroid. $ \\ $ By ampere’s circuital law, $ \\ \oint{\overrightarrow{B}.\overrightarrow{dl}}={{\mu }_{0}}I \\ \Rightarrow BL={{\mu }_{0}}I \\ $  Here, $L$ is the length of the toroid $ \\ L = 2\pi r \\ $ So, $BL={{\mu }_{0}}I \\ \Rightarrow B\left( 2\pi r \right)={{\mu }_{0}}NI \\ \therefore B=\dfrac{{{\mu }_{0}}NI}{2\pi r}$ $ \\ $ But inside the open space of the centroid, current enclosed is zero. So, $ \\ B=\dfrac{{{\mu }_{0}}N\times 0}{2\pi r} \\ \therefore B=0 \\ $ Therefore, the charge will not have any effect due to the magnetic field.

 A current of $2\,A$ is passed through a bangle made of copper. When measured using a thread, its circumference is found to be $12.48\,cm $. Then, what will be the magnetic field produced at the center of this bangle?  

 The expression for finding the magnetic field of a circular loop by Biot-savart’s law can be used to find the answer. Radius can be found using circumference and take $x=0$ as the point is at the center of the bangle.

 According to Biot-savart’s law, the magnetic field at point $x$ due to a coil is given by, $ \\ \overrightarrow{B}=\dfrac{{{\mu }_{0}}I}{4\pi }\dfrac{2\pi {{R}^{2}}}{{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}} \\ $ Here, $x=0$ as we need to find the field at the center of the bangle. So, $ \\ \overrightarrow{B}=\dfrac{{{\mu }_{0}}I}{4\pi }\dfrac{2\pi {{R}^{2}}}{{{\left( 0+{{R}^{2}} \right)}^{\dfrac{3}{2}}}} \\ \Rightarrow \overrightarrow{B}=\dfrac{{{\mu }_{0}}I}{4\pi }\dfrac{2\pi {{R}^{2}}}{\left( {{R}^{3}} \right)} \\ \Rightarrow \overrightarrow{B}=\dfrac{{{\mu }_{0}}I}{4\pi }\dfrac{2\pi {{R}^{2}}}{\left( {{R}^{3}} \right)} \\ \Rightarrow \overrightarrow{B}=\dfrac{{{\mu }_{0}}I}{2}\dfrac{1}{R} \\ \therefore \overrightarrow{B}=\dfrac{{{\mu }_{0}}I}{2R} \\ $ Now, circumference is given as 12.48 cm. then radius will be, $ \\ 2\pi R=12.48 \\ \therefore R=2\,cm \\ $ Now, Magnetic field due to the bangle will be, $ \\ \overrightarrow{B}=\dfrac{{{\mu }_{0}}I}{2R} \\ \Rightarrow \overrightarrow{B}=\dfrac{1.26\times {{10}^{-6}}\times 2}{2\times 2\times {{10}^{-2}}} \\ \therefore \overrightarrow{B}=6.3\times {{10}^{-5}}\,T $