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Properties of Charges

A charge part...

Question

A charge particle of mass 'm' & charge - q is released from rest from a given position Q inside a fixed vertical semicircular trough. For the presence and the absence of magnetic field normal reaction acting on the change at P are $N_{1} a n d N_{2}$ respectively. Neglecting friction and assuming gravity presence. The value of $N_{1} - N_{2}$ is:

2 B q \sqrt{2 g R}

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\frac{B q \sqrt{2 g R}}{2}

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B q \sqrt{2 g R}

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\frac{B q \sqrt{2 g R}}{4}

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Solution

The correct option is A $B q \sqrt{2 g R}$

Charge of particles of mass $= m$
charge $= - q$
The value of $B = \frac{μ_{0}}{4 π} \frac{I d l sin θ}{r^{2}}$ .
Now, the velocity across the field normal reaction acting on the change $= V = \sqrt{2 g R}$ . hence the friction difficence
$N_{1} - N_{2} = B q \sqrt{2 g R}$

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The correct option is A Bq√2gRCharge of particles of mass =mcharge =−qThe value of B=μ04πIdlsinθr2.Now, the velocity across the field normal reaction acting on the change =V=√2gR. hence the friction difficenceN1−N2=Bq√2gR