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A small block of mass \(20 \mathrm{~g}\) and charge \(4 \mathrm{mC}\) is released on a long smooth inclined plane of inclination angle of \(45^{\circ}\), A uniform horizontal magnetic field of \(1 \mathrm{~T}\) is acting parallel to the surface, as shown in the figure. The time from the start when the block loses contact with the surface of the plane is
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Verified Answer
The correct answer is:
\(5 \mathrm{~s}\)
Given inclination of plane,
Mass of block, \(m=20 \mathrm{~g}=0.02 \mathrm{~kg}\)
charge on the block, \(q=4 \mathrm{mC}=4 \times 10^{-3} \mathrm{C}\)
Magnetic field, \(B=1 \mathrm{~T}\)
Magnetic force on the charge particles, \(F=B q v\)
Particle will leave the inclined plane, when
\(F=m g \cos \theta \Rightarrow B q v=m g \cos \theta \Rightarrow v=\frac{m g \cos \theta}{q B}\)
Time taken to reached at the velocity \(v\) is given by
\(\begin{aligned}
& v=0+g \sin \theta t \quad[\because u=0, a=g \sin \theta] \\
& t=\frac{v}{g \sin \theta}=\frac{m g \cos \theta}{q B \cdot g \sin \theta}=\frac{m \cot \theta}{q B} \\
& t=\frac{0.02 \cot 45^{\circ}}{4 \times 10^{-3} \times 1}=5 \mathrm{~s} \quad\left[\because \cot 45^{\circ}=1\right]
\end{aligned}\)
Mass of block, \(m=20 \mathrm{~g}=0.02 \mathrm{~kg}\)
charge on the block, \(q=4 \mathrm{mC}=4 \times 10^{-3} \mathrm{C}\)
Magnetic field, \(B=1 \mathrm{~T}\)
Magnetic force on the charge particles, \(F=B q v\)
Particle will leave the inclined plane, when
\(F=m g \cos \theta \Rightarrow B q v=m g \cos \theta \Rightarrow v=\frac{m g \cos \theta}{q B}\)
Time taken to reached at the velocity \(v\) is given by
\(\begin{aligned}
& v=0+g \sin \theta t \quad[\because u=0, a=g \sin \theta] \\
& t=\frac{v}{g \sin \theta}=\frac{m g \cos \theta}{q B \cdot g \sin \theta}=\frac{m \cot \theta}{q B} \\
& t=\frac{0.02 \cot 45^{\circ}}{4 \times 10^{-3} \times 1}=5 \mathrm{~s} \quad\left[\because \cot 45^{\circ}=1\right]
\end{aligned}\)
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