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  • An electron gun is placed inside a long solenoid of radius R on its axis. The solenoid has n turns / length and carries a current I. The electron gun shoots an electron along the radius of the solenoid with speed <img alt="\nu" src="https://learn.car

An electron gun is placed inside a long solenoid of radius R on its axis. The solenoid has n turns / length and carries a current I. The electron gun shoots an electron along the radius of the solenoid with speed \nu . If the electron does not hit the surface of the solenoid, maximum possible value of \nu is (all symbols have their standard meaning) :  
Option: 1 \frac{2e\mu _{0}nIR}{m}      
      
Option: 2 \frac{e\mu _{0}nIR}{2m}

Option: 3 \frac{e\mu _{0}nIR}{4m}

Option: 4 \frac{e\mu _{0}nIR}{m}
 

Answers (2)

best_answer

 

Charge = e 

 

R_{max}=\frac{R}{2}=\frac{mv_{max}}{e\mu _{0}In}

Therefore, v_{max}=\frac{e\mu _{0}nIR}{2m}

R_{max}=\frac{R}{2}=\frac{mv_{max}}{e\mu _{0}In} 

Hence the correct option is (2).

Posted by

avinash.dongre

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Malus's Law -

Malus's Law-

This law states that the intensity of the polarized light transmitted through the analyzer varies as the square of the cosine of the angle between the plane of transmission of the analyzer and the plane of the polarizer.

                                       

As, 

                                                    I=I_{0} \cos ^{2} \theta

 

output intensity is given by I=I_0Cos^2(\theta )

Initial output intensity=10% of I_0

I.e \frac{10I_0}{100}=I_0Cos^2(\theta )\Rightarrow \theta =71.57

Final output intensity=O

means new angle is 90^0

the angle by which the analyser needs to be rotated further is 90^0-\theta =18.4^0

 

So Option (3) is correct.

Posted by

Ritika Jonwal

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