There is a constant homogeneous electric field of 100Vm –1 within the region x = 0 and x = 0.167 m pointing in the positive x-direction. There is a constant homogeneous magnetic field B within the region x = 0.167 m and x = 0.334m pointing in the z-direction. A proton at rest at the origin (x = 0, y = 0) is released in the positive x-direction. The minimum strength of the magnetic field B, so that the proton will come back at x = 0, y = 0.167 m (mass of the proton= 1.67 × 10 –27 kg) is...........mT.
Text Solution
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Sol. The situation described in the problem is shown in fig As electric field is along x-axis, so proton will be accelerated by the electric field and will enter the magnetic field at A(i.e., x = 0.167,
y = 0) with velocity v along x-axis such that
mv 2 = W = Fd = qEd
i.e. v = 
= 
= 4 
× 10 4 
Now as proton is moving perpendicular to magnetic field so it will describe a circular path in the magnetic field with radius r such that
r = 
And as it comes back at C[x = 0; y = 0.167m] its path in the magnetic field will be a semicircle such that
y = 2r = 
i.e. B = 
i.e., B = 
= 
× 10 –2 = 7.07 mT
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