JEE MAIN

<div><div><li> <p>Given below are two statements. One is as Assertion (A) and the other is labelled as Reason (R).</p> </li></div><p> <b>Assertion (A):</b> A electron in a certain region of uniform magnetic field is moving with constant velocity in a straight line path.<br><b>Reason (R):</b> The magnetic field in that region is along the direction of velocity of the electron.</p><p>In the light of the above statements, choose the correct answer from the options given below:</p></div>

<div><p> Force on electron in a uniform magnetic field will be:</p><p style="text-align:left;"> 
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     F = q(v \times B) 
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 </math> </p><p>If Assertion (A) is true then magnetic field must be along the velocity and force F is zero and hence v will be constant.</p><p>Thus A is correct and R is correct explanation</p></div>

<div><div><li> <p>Consider a moving coil galvanometer (MCG):<br> A. The torsional constant in moving coil galvanometer has dimensions [ML<sup>2</sup>T<sup>-2</sup>] B. Increasing the current sensitivity may not necessarily increase the voltage sensitivity.<br> C. If we increase number of turns (N) to its double (2N), then the voltage sensitivity doubles.<br> D. MCG can be converted into an ammeter by introducing a shunt resistance of large value in parallel with galvanometer.<br> E. Current sensitivity of MCG depends inversely on number of turns of coil.</p> </li></div><p>Choose the correct answer from the options given below:</p></div>

<div><p style="text-align:left;">Dimensionally, torsional constant: 
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     \tau = C\theta 
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     \lbrack C\rbrack = \lbrack\tau\rbrack = \left\lbrack {ML}^{2}{\text{\ }T}^{- 2} \right\rbrack 
   </annotation> 
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 </math> </p><p style="text-align:left;">if we try to increase current sensitivity (for instance, by increasing the number of turns or enhancing the magnetic field), the internal resistance may also change. 
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     Current\ sensitivity:\ S_{i} = \frac{\alpha}{i} = \frac{NBA}{C} 
   </annotation> 
  </semantics> 
 </math></p><p>Therefore, the improvement in current sensitivity does not automatically translate to a proportional increase in voltage sensitivity.</p><p>If we increase number of turns (N) to its double (2N), then resistance will also increase and Voltage sensitivity may NOT double.</p><p>MCG can be converted into an ammeter by introducing a shunt resistance of large value in SERIES with galvanometer</p><p>Current sensitivity of MCG depends DIRECTLY on number of turns of coil</p></div>

<div><div><li> <p>A bar magnet has total length 2l=20 units and the field point P is at a distance d=10 units from the centre of the magnet. If the relative uncertainty of length measurement is 1 %, then uncertainty of the magnetic field at point P is:</p> </li></div><p> <img src="http://test-master-question-data.s3.ap-south-1.amazonaws.com/questionImg/85d10323-5a61-41f2-89fa-9d2a93e58f6d.jpg" style="width:1.15229in;height:1.20672in" alt=""> </p></div>

<div><p style="text-align:left;"> 
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     Using\ Magnetic\ field:\ B = \frac{\mu_{0}M}{4{nr}^{3}}\ 
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     and\ dipole\ moment:\ \overrightarrow{M} = m\left( 2\overrightarrow{l} \right),\ 
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     where\ m\ is\ pole\ strength 
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 </math> </p><p style="text-align:left;"> 
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   <annotation encoding="application/x-tex">
     B = \frac{\mu_{0}m\ 2l}{4{nr}^{3}} 
   </annotation> 
  </semantics> 
 </math> </p></div>

<div><div><li> <p>A long straight wire of a circular cross-section with radius a carries a steady current I. The current I is uniformly distributed across this cross-section. The plot of magnitude of magnetic field B with distance r from the centre of the wire is given by –</p> </li></div></div>

<div><p>inside the wire, if we consider and Amperian loop:</p><p> <img src="http://test-master-question-data.s3.ap-south-1.amazonaws.com/questionImg/aba35e32-c1a3-4b95-b9ff-9112fbe73696.jpg" style="width:0.48192in;height:1.24723in" alt=""> </p><p style="text-align:left;"> 
 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <semantics> 
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       inc&nbsp; 
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   <annotation encoding="application/x-tex">
     B(2\pi r) = \mu_{0}I_{\text{inc~}} 
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  </semantics> 
 </math> </p><p style="text-align:left;"> 
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     \left\lbrack I_{\text{inc~}} = \frac{I}{\pi a^{2}} \times \pi r^{2} = \frac{I}{a^{2}} \cdot r^{2} \right\rbrack 
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  </semantics> 
 </math> </p><p style="text-align:left;"> <img src="http://test-master-question-data.s3.ap-south-1.amazonaws.com/questionImg/5855c60f-506e-4676-ae67-601109cac678.jpg" style="width:0.65083in;height:0.91266in" alt=""> 
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     B(2\pi r) = \mu_{0}\frac{I}{a^{2}}r^{2} 
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 </math> </p><p style="text-align:left;"> <img src="http://test-master-question-data.s3.ap-south-1.amazonaws.com/questionImg/ed1b6697-9965-4a63-a33c-f9568cec3ba0.jpg" style="width:1.22641in;height:0.93873in" alt=""> 
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     B(2\pi r) = \mu_{0}I 
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 </math> </p></div>

<div><div><li> <p>An infinite wire has a circular bend of radius a, and carrying a current I as shown in figure. The magnitude of magnetic field at the origin O of the arc is given by:</p> </li></div><p> <img src="http://test-master-question-data.s3.ap-south-1.amazonaws.com/questionImg/f04af695-9a9c-4905-8957-e77f17bf01c8.jpg" style="width:1.58316in;height:0.92127in" alt=""> </p></div>

<div><p>Net magnetic field at O due to three wire segments</p><p style="text-align:left;"> <img src="http://test-master-question-data.s3.ap-south-1.amazonaws.com/questionImg/053b4a21-848c-4d80-9b4f-a09a9faae64f.jpg" style="width:2.1692in;height:1.01312in" alt=""> 
 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <semantics> 
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   <annotation encoding="application/x-tex">
     B = B_{1} + B_{2} + B_{2} 
   </annotation> 
  </semantics> 
 </math> </p><p style="text-align:left;"> 
 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <semantics> 
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   <annotation encoding="application/x-tex">
     B_{1} = 0 
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 </math> </p><p style="text-align:left;"> 
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     B_{2} = \frac{\mu_{0}i}{4\pi a}\theta = \frac{\mu_{0}i}{4\pi a} \times \frac{3\pi}{2} 
   </annotation> 
  </semantics> 
 </math> </p><p style="text-align:left;"> 
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   </mrow> 
   <annotation encoding="application/x-tex">
     B_{3} = \frac{\mu_{0}i}{4\pi a} 
   </annotation> 
  </semantics> 
 </math> </p><p style="text-align:left;"> 
 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <semantics> 
   <mrow> 
    <mi>
      B 
    </mi> 
    <mo>
      = 
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    <mfrac> 
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         0 
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        i 
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        4 
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        π 
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       [ 
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       + 
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         3 
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        2 
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     <mo stretchy="true" form="postfix">
       ] 
     </mo> 
    </mrow> 
   </mrow> 
   <annotation encoding="application/x-tex">
     B = \frac{\mu_{0}i}{4\pi a}\left\lbrack 1 + \frac{3\pi}{2} \right\rbrack 
   </annotation> 
  </semantics> 
 </math> </p></div>

<div><div><li> <p>Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire's cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, anywhere due to the wire, will be:</p> </li></div></div>

<div><p style="text-align:left;">Maximum possible magnetic field is at the surface<br> It can be obtained inside as well as outside the wire for inside,<br> 
 <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"> <semantics> 
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   <annotation encoding="application/x-tex">
     \frac{\mu_{0}I}{4\pi a} = \frac{\mu_{0}Ir}{2\pi a^{2}} 
   </annotation> 
  </semantics> 
 </math> 
 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <semantics> 
   <mrow> 
    <mi>
      U 
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      s 
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      g 
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    <mspace width="0.222em"></mspace> 
    <msub> 
     <mi>
       B 
     </mi> 
     <mtext mathvariant="normal">
       Surface 
     </mtext> 
    </msub> 
    <mo>
      = 
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     <mi>
       B 
     </mi> 
     <mtext mathvariant="normal">
       max 
     </mtext> 
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      = 
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        a 
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   <annotation encoding="application/x-tex">
     Using\ B_{\text{Surface}} = B_{\text{max}} = \frac{\mu_{0}I}{2\pi a} 
   </annotation> 
  </semantics> 
 </math></p><p style="text-align:left;">for outside<br> 
 <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"> <semantics> 
   <mrow></mrow> 
   <annotation encoding="application/x-tex"></annotation> 
  </semantics> 
 </math> 
 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <semantics> 
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        r 
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   </mrow> 
   <annotation encoding="application/x-tex">
     \frac{\mu_{0}I}{4\pi a} = \frac{\mu_{0}Ir}{2\pi a^{2}} 
   </annotation> 
  </semantics> 
 </math> 
 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <semantics> 
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      r 
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       2 
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   </mrow> 
   <annotation encoding="application/x-tex">
     r = \frac{a}{2} 
   </annotation> 
  </semantics> 
 </math></p><p style="text-align:left;">Correct answer 
 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <semantics> 
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   <annotation encoding="application/x-tex">
     r = 2a 
   </annotation> 
  </semantics> 
 </math> 
 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <semantics> 
   <mrow> 
    <mo stretchy="true" form="prefix">
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   <annotation encoding="application/x-tex">
     \left\lbrack \frac{a}{2},2a \right\rbrack 
   </annotation> 
  </semantics> 
 </math></p></div>

<div><div><li> <p>N equally spaced charges each of value q, are placed on a circle of radius R. The circle rotates about its axis with an angular velocity ω as shown in the figure. A bigger Amperian loop B encloses the whole circle where as a smaller Amperian loop A encloses a small segment. The difference between enclosed currents, I<sub>A</sub>-I<sub>B</sub>, for the given Amperian loops is -</p> </li></div><p> <img src="http://test-master-question-data.s3.ap-south-1.amazonaws.com/questionImg/9c005199-1809-491e-b43e-39e46943d24f.jpg" style="width:1.47717in;height:1.05087in" alt=""> </p></div>

<div><p style="text-align:left;">For loop 'A'<br> For loop 'B'<br> (because net incoming or outgoing charge will cancel each other)<br> 
 <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"> <semantics> 
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     </mrow> 
    </mfrac> 
   </mrow> 
   <annotation encoding="application/x-tex">
     I_{A} - I_{B} = \frac{N^{2}q\omega}{2\pi} 
   </annotation> 
  </semantics> 
 </math> 
 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <semantics> 
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   </mrow> 
   <annotation encoding="application/x-tex">
     I_{A} = \frac{q_{mt}}{T} = \frac{Nq}{\left( \frac{2\pi}{\omega} \right)} = \frac{Nq\omega}{2\pi} 
   </annotation> 
  </semantics> 
 </math> 
 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <semantics> 
   <mrow> 
    <msub> 
     <mi>
       I 
     </mi> 
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    <mo>
      = 
    </mo> 
    <mn>
      0 
    </mn> 
    <mrow> 
     <mspace width="0.333em"></mspace> 
     <mtext mathvariant="normal"></mtext> 
     <mspace width="0.333em"></mspace> 
    </mrow> 
   </mrow> 
   <annotation encoding="application/x-tex">
     I_{B} = 0\text{\ } 
   </annotation> 
  </semantics> 
 </math></p></div>

<div><div><li> <p>Consider a long thin conducting wire carrying a uniform current I. A particle having mass M and charge q is released at a distance a from the wire with a speed v<sub>o</sub> along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance x from the wire. The value of x is: (μ<sub>0</sub> is vacuum permeability)</p> </li></div></div>