MHT CET · Physics · Magnetic Effects of Current
A proton moving in perpendicular magnetic field possess energy \(E\). The strength of magnetic field is increases four times. But the proton is constrained to move in the path of same radius. The kinetic energy of proton will increase
- A 4 times
- B 12 times
- C 8 times
- D 16 times
Answer & Solution
Correct Answer
(A) 4 times
Step-by-step Solution
Detailed explanation
In orthogonal magnetic field, the changed particle undergoes uniform circular motion.
\(\begin{aligned} & \therefore \frac{m v^2}{R}=q v B \\ & \Rightarrow v^2\left(\frac{q B}{m}\right) r\end{aligned}\)
If B increases by four times \(\& r\) remains same
\(\begin{aligned} & v_{\text {new }}^2=\left[\left(\frac{q B}{m}\right) r\right] 4=(2 v)^2 \\ & \therefore K E_{\text {new }}=\frac{1}{2} v_{\text {new }}^2=\left(\frac{1}{2} m v^2\right) 4\end{aligned}\)
\(\begin{aligned} & \therefore \frac{m v^2}{R}=q v B \\ & \Rightarrow v^2\left(\frac{q B}{m}\right) r\end{aligned}\)
If B increases by four times \(\& r\) remains same
\(\begin{aligned} & v_{\text {new }}^2=\left[\left(\frac{q B}{m}\right) r\right] 4=(2 v)^2 \\ & \therefore K E_{\text {new }}=\frac{1}{2} v_{\text {new }}^2=\left(\frac{1}{2} m v^2\right) 4\end{aligned}\)
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